maths teacher support: - fet euclidean geometry

MATHS TEACHER SUPPORT:FET Euclidean Geometry: The Power of Proof

25 January 2022

Hosted by Gretel Lampe

Presented by Anne Eadie

Q & A by Jenny Campbell & Susan Carletti

2022 Maths Teacher

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Mathematics

Euclidean Geometry:Power of Proof

1

The Major Issues

Language

Knowledge

Logic

and then

Strategies

SUMMARY OF

CONTENTS OF BOOKLETS

BOOKLET 1: Teaching Documents

BOOKLET 2: Lines, Angles, Triangles & Quadrilaterals

BOOKLET 3: Circle Geometry

BOOKLET 4: Proportionality, Similarity and

The Theorem of Pythagoras

These Geometry materials (Booklets 1 to 4) were created and produced by The Answer Series Educational Publishers (Pty) (Ltd)

to support the teaching and learning of Geometry in high schools in South Africa.

They are freely available to anyone who wishes to use them.

This material may not be sold (via any channel) or used for

profit-making of any kind.

https://www.theanswer.co.za/course/teacher-development-euclidean-geometry/

WWW.THEANSWER.CO.ZA

Teaching Documents

by The TAS Maths Team

TAS FET EUCLIDEAN GEOMETRY COURSE

BOOKLET 1

Mastering FET Geometry

Geometry FET Course Booklets Set

Booklet 1: TEACHING DOCUMENTS

Exam Mark Distribution (FET Maths Paper 2)

The 2022 ATP (Proposed)

Research: Results & Diagnostic Reports (2020)

Exemplar FET Geometry Questions and Detailed Solutions

Problem-solving approach: ACT

The CAPS Curriculum Overview (FET Geometry)

The FET Geometry CAPS Curriculum per Grade (10 – 12)

Acceptable reasons for statements in Geometry

Important Advice for Learners

FET EXAM: Mark distribution

PAPER 2 Proofs: maximum 12 marks

Description GR 10 GR 11 GR 12

Statistics 15 20 20

Analytical Geometry 15 30 40

Trigonometry 40 50 50

Euclidean Geometry & Measurement 30 50 40

TOTAL 100 150 150

NOTE:

• Questions will not necessarily be compartmentalised in sections, as this table indicates. Various topics can be integrated in the same question.

• A formula sheet will be provided for the final examinations in Grades 10, 11 and 12.

CAPS Curriculum page 17

January 2022

A PROPOSED 2022 ATP FOR FET MATHS

Grade 10 Grade 11 Grade 12

No. of

weeks

No.

of weeks

No.

of weeks

Te

rm

1

Algebraic Expressions,

Numbers & Surds Exponents, Equations & Inequalities Equations & Inequalities

Euclidean Geometry (#1)

4

2

1

3

Exponents & Surds

Equations

Equations & Inequalities

Euclidean Geometry

Trig functions &

Revision of Grade 10 Trig

Trig identities &

Reduction formulae

1

1

2

4

1

1

Patterns, Sequences and Series

Euclidean Geometry

Trigonometry

(Algebra)

3

3

4

Te

rm

2 Trigonometry (#1)

Number Patterns

Functions (including

Trig Functions (#2))

Measurement

3

1

6

2

Trig equation & General solutions

Quadrilaterals

Analytical Geometry

Number Patterns

Functions

Trig – sin/cos/area rules

1

1

2

2

5

1

Analytical Geometry Functions & Inverse Functions

& Exp & Log Functions Calculus, including Polynomials Finance

2

2

5

3

Te

rm

3 Statistics

Probability

Finance (Growth)

Analytical Geometry

Euclidean Geometry (#2)

2

2

2

2

3

Trig – sin/cos/area rules

Measurement

Statistics

Probability

Finance (Growth & Decay)

1

2

3

4

1

Finance

Statistics (regression & correlation)

Counting and Probability

INTERNAL EXAMS

1

3

3

4

Te

rm

4 Revision

FINAL EXAMS

Reporting

4.

3

1 ½

Finance (Growth & Decay)

Revision

FINAL EXAMS

Reporting

3

1

4

1½

Revision (Paper 1)

Revision (Paper 2)

Revision (Exam Techniques?)

EXTERNAL EXAMS

1

1

1

6½

RESEARCH:Results & Diagnostic Reports (2020)

Some interesting statistics ...

PAPER 2:

ave. over 7 years

2014 – 2020 2020 2021?

Statistics 60,1% 74,4%

Analytical Geometry 54,3% 52,3%

Trigonometry 41,6% 46,7%

Euclidean Geometry

46,2% 42,9%

Paper 2 50,6% 50,6%

vs Paper 1 52,2% 50,6%

Note: This information is based on a random sample of candidates and may not reflect

the national averages accurately.

However, it is useful in assessing the RELATIVE degrees of each question

AS EXPERIENCED BY CANDIDATES.

2020: Paper 2

Average % performance per question

Q1 Data Handling

Q2 Data Handling

Q3 Analytical Geometry

Q4 Analytical Geometry

Q5 Trigonometry

Q6 Trigonometry

Q7 Trigonometry

Q8 Euclidean Geometry



and, per sub-question

Ave

rage

per

form

ance

(%)

Sub-question

Copyright © The Answer Series 1

DIAGNOSTIC REPORT: DBE NOV 2020

QUESTION

8.1 O is the centre of the circle.

KOM bisects chord LN and MNOˆ = 26º.

K and P are points on the circle with

NKPˆ = 32º. OP is drawn.

8.1.1 Determine, giving reasons,

the size of:

(a) 2

Oˆ (b) 1

Oˆ (2)(4)

8.1.2 Prove, giving reasons, that

KN bisects OKP.ˆ (3)

Common Errors and Misconceptions

(a) In Q8.1.1 (a) some candidates assumed that ˆ2

Q = ˆ1

K because these

angles appeared to be in the same segment. It was incorrect to assume

that KPNO was a cyclic quadrilateral.

(b) Some candidates incorrectly assumed that ˆ2

M = 90°. Other candidates

were unable to provide the correct reason for the statement ˆ2

M = 90°.

They wrote ‘line from centre to midpoint’ or just ‘perpendicular lines’.

Neither of these was accepted.

(c) Some candidates made the following incorrect assumptions when

answering Q8.1.2:

ˆ

2K = ˆ

1K (as if it was given information), ON || KP and OP ⊥. KN.

Some candidates incorrectly stated that ˆP = ˆ1

O . These candidates

assumed that these angles were opposite sides of equal length.

1 2

3

K O

26º

P

1 2

1

2

L

M

N

32º


QUESTION

8.1 O is the centre of the circle. KOM bisects chord LN and

MNOˆ = 26º. K and P are points on the circle with NKPˆ = 32º.

OP is drawn.

8.1.1 Determine, giving reasons, the size of :

(a) 2

Oˆ (2)

(b) 1

Oˆ (4)

8.1.2 Prove, giving reasons, that KN bisects OKP.ˆ (3)

MEMO

8.1

8.1.1 (a) ˆ

2O = 2(32º) . . .

= 64º �

(b) ˆ

1O + ˆ

2O = 26º + ˆ 2M . . .

& ˆ 2M = 90º . . .

∴ ˆ

1O + 64º = 26º + 90º

∴ ˆ

1O = 52º �

OR: ˆ

2M = 90º . . .

∴ ˆ 3O = 64º . . . ø sum in Δ

∴ ˆ

1O = 52º � . . . øs

on a straight line

8.1.2 ˆ

2K = ˆKNO . . . øs

opp equal radii

= ˆ

3

1O

2 . . . exterior ø of Δ

= 32º �

KN bisects ˆOKP �

ø at centre = 2 % ø at circumference

exterior ø of Δ

line from centre to midpoint of chord

line from centre to midpoint of chord

1 2

3

K O

26º

P

1 2

1

2

L

M

N

32º

1 2

3

K O

26º

P

1 2

1

2

L

M

N

32º


QUESTION

8.2 In ΔABC, F and G are points on sides AB

and AC respectively.

D is a point on GC such that 1

Dˆ = Bˆ .

8.2.1 If AF is a tangent to the circle

passing through points F, G and D,

then prove, giving reasons, that

FG || BC. (4)

8.2.2 If it is further given that AF 2

FB 5 = ,

AC = 2x – 6 and GC = x + 9,

then calculate the value of x. (4)

[17]


(d) In Q8.2.1 many candidates made the incorrect assumption that FG || BC.

Using this information, they would conclude that ˆB = ˆ1F because the

corresponding angles were equal. Other candidates made unnecessary

constructions and wanted to prove the proportionality theorem.

Some candidates could not provide a valid reason as to why FG was

parallel to BC.

(e) Candidates were unable to state the correct proportion in Q8.2.2.

Those who could identify the correct proportion were unable to

simplify correctly.

7x + 63 = 10x – 30

–7x + 10x = –30 + 63

Candidates failed to provide a reason for the proportion that they wrote.

Algebra!

A

C

F

G

B

D

1 2

1

2


QUESTION

8.2 In ΔABC, F and G are points on sides AB and AC respectively.

D is a point on GC such that 1

Dˆ = B.ˆ

8.2.1 If AF is a tangent to the circle passing through

points F, G and D, then prove, giving reasons,

that FG || BC. (4)

8.2.2 If it is further given that AF 2 =

FB 5, AC = 2x – 6

and GC = x + 9, then calculate the value of x. (4) [17]

MEMO

8.2.1 ˆ

1F = ˆ 1D . . . tan chord theorem

= ˆB . . . given

∴ FG | | BC . . . corresp. øs

:

ˆ ˆ

1F = B

8.2.2 AG = (2x – 6) – (x + 9)

= x – 15

AC

GC =

AB

FB . . . proportion theorem FG | | BC

∴ – 15

+ 9

x

x

= 2

5

∴ 5x – 75 = 2x + 18

∴ 3x = 93

∴ x = 31 �

OR: AB : FB = 7 : 5

AC

GC =

AB

FB . . . proportion theorem FG | | BC

∴ –2 6

+ 9

x

x

= 7

5

∴ 10x – 30 = 7x + 63

∴ 3x = 93

∴ x = 31 �

A converse theorem

A

C

F G

B

D

2

1

2

1

B

A

C

F G

x – 15

x + 9 5

2

B

A

C

F G

x + 9

2x – 62

7

5

A

C

F

G

B

D

1 2

1

2


QUESTION 8: Suggestions for Improvement

(a) Learners should be encouraged to scrutinise the given information and the diagram for clues about which theorems

could be used in answering the question.

(b) Teachers must cover the basic work thoroughly. An explanation of the theorem should be accompanied by

showing the relationship in a diagram.

(c) Learners should be told not to make assumptions based on what they see in the diagram. They should be reminded

that the diagrams are not drawn to scale.

(d) Learners should be taught that all statements must be accompanied by reasons. It is essential that the parallel lines

be mentioned when stating that corresponding angles are equal, alternate angles are equal, the sum of the

co-interior angles is 180° or when stating the proportional intercept theorem.

(e) Learners should know that writing a correct statement and reason does not guarantee marks.

They will only get marks if that statement and reason leads to the solution.


QUESTION

9.1 In the diagram, O is the centre of the circle.

Points S, T and R lie on the circle.

Chords ST, SR and TR are drawn in the circle.

QS is a tangent to the circle at S.

Use the diagram to prove the theorem which

states that QST = R.ˆ ˆ (5)


(a) In answering Q9.1, some candidates either did not do

the construction correctly or they failed to do the construction

altogether. Other candidates attempted to use more than

one method to prove this theorem but did not reach a

conclusion through any method.

T

O

R

Q

S


QUESTION

9.1 In the diagram, O is the centre of the circle.

Points S, T and R lie on the circle.

Chords ST, SR and TR are drawn in the circle.

QS is a tangent to the circle at S.

Use the diagram to prove the theorem which states that

QST = R.ˆ ˆ (5)

MEMO

9.1 The proof of the tan chord theorem:

Let ˆQST = x

ˆOSQ = 90º . . . radius ⊥ tangent

∴ ˆ 1S = 90º – x

∴ ˆ1T = 90º – x . . . øs

opp equal radii

∴ ˆTOS = 180º – 2(90º – x) . . . sum of øs

of Δ

= 2x

∴ ˆR = x . . . ø at centre = 2 % ø at circumference

∴ ˆQST = ˆR �

T

O

R

Q

S


QUESTION

9.2 Chord QN bisects MNPˆ and intersects chord MP at S.

The tangent at P meets MN produced at R such that

QN || PR. Let 1

Pˆ = x.

9.2.1 Determine the following angles in terms of x.

Give reasons.

(a) 2

Nˆ (2)

(b) 2

Qˆ (2)

9.2.2 Prove, giving reasons, that MN MS =

NR SQ (6)

[15]


(b) In Q9.2.1 many candidates did not provide a correct or

complete reason for their statements.

(c) Instead of using the proportionality theorem to answer Q9.2.2,

candidates attempted to use similar triangles to find the ratios.

Some incorrectly assumed that S was the centre of the circle

and that MNPQ was a square. From these assumptions, they

then stated that there were a number of sides that were equal

in length.

1 2

3

M

S

P

1 2

1

2

N

Q

x

2

3

1

R


QUESTION

9.2 Chord QN bisects MNPˆ and intersects chord MP at S.

The tangent at P meets MN produced at R such that QN || PR.

Let 1

Pˆ = x.

9.2.1 Determine the following angles in terms of x.

Give reasons.

(a) 2

Nˆ (2)

(b) 2

Qˆ (2)

9.2.2 Prove, giving reasons, that MN MS =

NR SQ (6)

[15]

MEMO

9.2.1 (a) ˆ

2N = ( )ˆ

1P = x � . . . alternate øs

; QN | | PR

(b) ˆ

2Q = ( )ˆ

1P = x � . . . tan chord theorem

9.2.2

In ΔMPR: MN

NR =

SP

MS . . . proportion theorem ; SN | | PR

ˆ

3P = ˆ

1N . . . øs

in same segment

= ˆ

2N . . . Given: QN bisects ˆMNP

= x

& ˆ

2Q = x . . . in Q9.2.1(b)

∴ ˆ3P = ˆ

2Q (= x)

∴ SP = SQ . . . sides opposite equal øs

∴ MN

NR =

MS

SQ �

R

1 2

3

M

S

P

1 2

1

2

Q

x

2

3

1

x

x

x

x

N

We need to prove that

SP = SQ

1 2

3

M

S

P

1 2

1

2

N

Q

x

2

3

1

R

Mark the sides MN, NR, MS & SQ on the sketch.

Three of these lengths and the | | lines point towards proportion theorem !



(a) Learners should be taught that a construction is required in order to prove a theorem. If the construction is

not shown, then the proof is regarded as a breakdown and they get no marks. Teachers should test theory

in short tests and assignments.

(b) Learners should be discouraged from writing correct statements that are not related to the solution.

No marks are awarded for statements that do not lead to solving the problem.

(c) Learners should be forced to use acceptable reasons in Euclidean Geometry. Teachers should explain

the difference between a theorem and its converse. They should also explain the conditions for which

theorems are applicable and when the converse will apply.

(d) Learners need to be told that success in answering Euclidean Geometry comes from regular practice, starting off

with the easy and progressing to the difficult.


QUESTION

QUESTION 10

In the diagram, a circle passes through

D, B and E. Diameter ED of the circle is produced to C

and AC is a tangent to the circle at B. M is a point on DE such that AM ⊥ DE. AM and chord BE intersect at F.

10.1 Prove, giving reasons, that: 10.1.1 FBDM is a cyclic quadrilateral (3)

10.1.2 3 1

B = Fˆ ˆ (4)

10.1.3 ΔCDB ||| ΔCBE (3)


(a) Some candidates assumed that FBDM was a cyclic quadrilateral instead

of proving that it was. Many candidates did not provide the correct reason

when concluding why FBDM was a cyclic quadrilateral.

A common error was ‘opp øs of a cyclic quad’.

(b) In Q10.1.2 some candidates did not use the knowledge that FBDM was

a cyclic quadrilateral. They incorrectly assumed that AM was a tangent

to the circle FBDM. Some candidates referred to the incorrect angle at

point B. They considered the entire ˆB instead of ˆ1

B ; ˆ2

B or ˆ3

B .

(c) When answering Q10.1.3, some candidates attempted to prove that the

triangles were congruent instead of trying to prove them similar.

Some candidates were able to identify the equal pairs of angles but

could not provide the correct reasons for them being equal. Other

candidates could not name the angles correctly. They would state that

ˆB = ˆE instead of ˆ1

B = ˆ1

E and that ˆD = ˆB instead of ˆ1

D = ˆCBE .

1

2

3

M

B

1 2 1 2

D

2 3

1

C

4

E

F

A


QUESTION

QUESTION 10

In the diagram, a circle passes through D, B and E. Diameter ED of the circle is produced to C and AC

is a tangent to the circle at B.

M is a point on DE such that AM ⊥ DE. AM and chord BE intersect at F.

10.1 Prove, giving reasons, that :

10.1.1 FBDM is a cyclic quadrilateral (3)

10.1.2 3 1B = Fˆ ˆ (4)

10.1.3 ΔCDB ||| ΔCBE (3)

MEMO

10.1.1

ˆ

2M = 90º

& ˆ 2B = 90º . . . ø in semi-?

∴ ˆ 2M = ˆ

2B

∴ FBDM is a cyclic quad. � . . . CONVERSE of ext. ø of cyclic quad.

10.1.2 ˆ

3B = ˆ 2D . . . tan chord theorem

= ˆ1F � . . . ext.ø of c.q. FBDM (10.1.1)

10.1.3

In Δs CBD & CBE

(1) ˆC is common

(2) ˆ

1B = ˆE . . . tan chord theorem

∴ ΔCDB | | | ΔCBE � . . . øøø

Shade the quadrilateral

Mark the Δs clearly

1

2

3

M

B

1 2 1 2

D

2 3

1

C

4

E

F

A

1

2

3

M

B

1 2 12

D

23

1

C

4

E

F

A


QUESTION

10.2* If it is further given that CD = 2 units and DE = 6 units,

calculate the length of:

10.2.1 BC (3)

10.2.2 DB (4) [17]

Misconceptions

(d) In Q10.2.1 many candidates could not establish the correct

proportion from Q10.1.3. They incorrectly assumed that

EC = DE and some candidates substituted BC = DE – CD

(e) Many candidates did not attempt Q10.2.2. Of those who did,

some were able to establish the correct proportion but could

not proceed any further.

1

2

3

M

B

1 2 1 2

D

2 3

1

C

4

E

F

A


QUESTION

10.2* If it is further given that CD = 2 units and DE = 6 units,

calculate the length of :

10.2.1 BC (3)

10.2.2 DB (4)

[17]

MEMO

10.2.1 ∴ DB

=CD BC

= BC

E BEC

⎛ ⎞⎜ ⎟⎝ ⎠

. . . | | |Δs

∴ 2

BC =

BC

2 + 6

∴ BC2 = 16

∴ BC = 4 units �

10.2.2 DB

BE =

CD

BC =

2

4 =

1

2

∴ BE = 2DB

Let DB = x, then BE = 2x

In ΔDBE: ˆ 2B = 90º . . . ø in semi-?

∴ DB2 + BE2 = DE2 . . . Theorem of Pythagoras

∴ x2 + (2x)2 = 62

∴ x2 + 4x2 = 36

∴ 5x2 = 36

∴ x2 = 36

5

∴ x 2,68 units �

1

2

3

M

B

1 2 1 2

D

2 3

1

C

4

E

F

A



(a) More time needs to be spent on the teaching of Euclidean Geometry in all grades. More practice on Grade 11

and 12 Euclidean geometry will help learners to learn theorems and diagram analysis. They should carefully read

the given information without making any assumptions. This work covered in class must include different activities

and all levels of taxonomy.

(b) Teachers should require learners to make use of the diagrams in the answer book to indicate angles and

sides that are equal and record information that has been calculated.

(c) Learners need to be made aware that writing correct but irrelevant statements will not earn them any marks

in an examination.

(d) Learners need to be exposed to questions in Euclidean Geometry that include the theorems and the converses.

When proving that a quadrilateral is cyclic, no circle terminology may be used when referring to the quadrilateral.


GR 10 – 12 EXEMPLAR GEOMETRY

GRADE 10: QUESTIONS

1. PQRS is a kite such that the diagonals intersect in O.

OS = 2 cm and ˆOPS = 20º.

1.1 Write down the length of OQ. (2)

1.2 Write down the size of ˆPOQ. (2)

1.3 Write down the size of ˆQPS. (2) [6]

2. In the diagram, BCDE and AODE are parallelograms. 2.1 Prove that OF || AB. (4) 2.2 Prove that ABOE is a parallelogram. (4)

2.3 Prove that ABO ≡ EOD. (5) [13]

GRADE 10: MEMOS

1.1 OQ = 2 cm � . . .

1.2 ˆPOQ = 90º � . . .

1.3 ˆQPO = 20º . . .

â ˆQPS = 40º �

2.

2.1 In DBA:

O is the midpt of BD . . .

& F is the midpt of AD . . .

â OF || AB � . . .

2.2 AE || OD . . . opp. sides of ||m AODE

â AE || BO

and OF || AB . . . proven above

â OE || AB

â ABOE is a ||m . . .

OR: In ||m AODE: AE = and || OD . . .

But OD = BO . . .

â AE = and || BO

â ABOE is a ||m � . . .

2.3 In s ABO and EOD

1) AB = EO . . . opposite sides of ||m ABOE

2) BO = OD . . . proved in 2.1

3) AO = ED . . . opposite sides of ||m AODE

â ABO ≡ EOD � . . . SSS

the longer diagonal of a kite

bisects the shorter diagonal

the diagonals of a kite

intersect at right angles

the longer diagonal of a

kite bisects the (opposite)

angles of a kite

Hint:

Use highlighters to mark the various ||ms and s

diagonals of ||m BCDE

bisect each other

diagonals of ||m AODE

bisect each other

the line joining the

midpoints of two sides

of a is || to the 3rd side

both pairs of opposite

sides are parallel

opp. sides

of ||m

O proved midpt of BD

of BD in 2.1

1 pr of opp. sides

= and || A

B

C D

E

O

F The highlighted s

(and their sides)

refer to Question 2.3.

A

B

C D

E

O

F

Q

P

S

RO

2 cm

20º


GRADE 11: QUESTIONS

1.1 Complete the statement so that it is valid:

The line drawn from the centre of the circle

perpendicular to the chord . . . (1)

1.2 In the diagram, O is the

centre of the circle. The diameter DE is

perpendicular to the

chord PQ at C. DE = 20 cm and CE = 2 cm.

Calculate the length of the following with reasons: 1.2.1 OC 1.2.2 PQ (2)(4) [7]

2.1 In the diagram, O is the

centre of the circle and

A, B and D are points on

the circle.

Use Euclidean geometry methods to prove the

theorem which states that ˆAOB = ˆ2ADB. (5)

2.2 In the diagram, M is the centre of the circle.

A, B, C, K and T lie on the circle. AT produced and CK produced meet in N.

Also NA = NC and ˆB = 38º.

2.2.1 Calculate, with reasons, the size of the

following angles:

(a) ˆKMA (b) ˆ

2T (2)(2)

(c) ˆC (d) ˆ

4K (2)(2)

2.2.2 Show that NK = NT. (2)

2.2.3 Prove that AMKN is a cyclic

quadrilateral. (3) [18]

3.1 Complete the following statement so that it

is valid: The angle between a chord and a tangent

at the point of contact is . . . (1)

3.2 In the diagram, EA is a tangent to circle ABCD

at A. AC is a tangent to circle CDFG at C. CE and AG intersect at D.

If ˆ 1A = x and ˆ 1E = y, prove the following with

reasons:

3.2.1 BCG || AE (5)

3.2.2 AE is a tangent to circle FED (5)

3.2.3 AB = AC (4) [15]

B

1 A

C

D

E

F

G

1

1

1

1

1

2

2

2

2

2

2

33

3

4

x

5

y

P

E

C

O

D

Q

D

O

A

B

N

T

38º

1

1 2

2

3

4

B

C M

A

K


GRADE 11: MEMOS

1.1 . . . bisects the chord �

1.2.1 OE = OD = 1

2(20) = 10 cm

â OC = 8 cm � . . . CE = 2 cm

1.2.2 In OPC:

PC2 = OP2 - OC2 . . . Pythagoras

= 102 - 82

= 36

â PC = 6 cm

â PQ = 12 cm � . . . line from centre chord

2.1 Construction: Join DO and produce it to C Proof :

Let ˆ1

D = x

then Â = x . . . â ˆ

1O = 2x

. . . ext. ø of DAO

Similarly: Let ˆ2

D = y

then, ˆ

2O = 2y

â ÂOB = 2x + 2y

= 2(x + y)

= 2 ÂDB �

2.2

2.2.1 (a) ˆKMA = 2(38º) . . .

= 76º �

(b) ˆ

2T = 38º � . . . ext. ø of cyclic quad. BKTA

(c) ˆC = 38º � . . .

(d) ˆNAC = 38º . . . øs opp = sides

â ˆ4

K = 38º � . . . ext. ø of c.q. CKTA

2.2.2 In NKT: ˆ

4K = ˆ

2T . . . both = 38º in 2.2.1

â NK = NT � . . . sides opp equal øs

2.2.3 ˆKMA = 2(38º) . . . see 2.2.1(a)

& ˆN = 180º - 2(38º) . . . sum of øs in NKT

(see 2.2.2)

â ˆKMA + ˆN = 180º

â AMKN is a cyclic quadrilateral �

. . . opposite øs supplementary

3.1 . . . equal to the angle subtended by the chord

in the alternate segment. �

3.2

3.2.1 ˆ

1A = x . . . given

â ˆ

2C = x . . . tan chord theorem

â ˆ

2G = x . . . tan chord theorem

â ˆ

1A = (alternate) ˆ

2G

â BCG || AE � . . . (alternate øs equal)

3.2.2 ˆ

1F = ˆ

3C . . . ext. ø of cyclic quad. CGFD

= ˆ

1E (= y) . . . alternate øs ; BCG || AE

â AE is a tangent to ?FED �

. . . converse of tan chord theorem

3.2.3 ˆ

1C = ˆCAE . . . alternate øs ; BCG || AE

= ˆB . . . tan chord theorem

â AB = AC � . . . sides opposite equal øs

radii

= 12

diameter

ø at centre =

2 % ø at circumference

øs in the same segment

or, ext. ø of cyclic quad. CKTA

P

E

C

O

D

Q

radii ;

øs opp = sides

B

1 A

C

D

E

F

G

1

1

1

1

1

2

2

2

2

2

2

33

3

4

x 5

y

D

O

A

B C

1

1 2

2

N

T

38º

1

12

2

3

4

B

C M

A

K


GRADE 12: QUESTIONS

1.1 Complete the following statement : The angle between the tangent and the chord at

the point of contact is equal to . . . (1)

1.2 In the diagram, A, B, C, D and E are points on the

circumference of the circle such that AE | | BC.

BE and CD produced meet in F. GBH is a

tangent to the circle at B. ˆ1B = 68º and ˆF = 20º.

Determine the size of each of the following:

1.2.1 ˆ

1E (2)

1.2.2 ˆ

3B (1)

1.2.3 ˆ

1D (2)

1.2.4 ˆ

2E (1)

1.2.5 ˆC (2) [9]

2. In the diagram, M is the centre of the circle and diameter

AB is produced to C. ME is drawn perpendicular to AC

such that CDE is a tangent to the circle at D. ME and

chord AD intersect at F. MB = 2BC.

2.1 If ˆ 4D = x, write down, with reasons, TWO other

angles each equal to x. (3)

2.2 Prove that CM is a tangent at M to the circle

passing through M, E and D. (4)

2.3 Prove that FMBD is a cyclic quadrilateral. (3)

2.4 Prove that DC2 = 5BC2. (3)

2.5 Prove that DBC | | | DFM. (4)

2.6 Hence, determine the value of .

DM

FM (2) [19]

3.1 In the diagram, points D

and E lie on sides AB

and AC respectively of

ABC such that DE | | BC.

Use Euclidean Geometry

methods to prove the

theorem which states

that .

AD AE

DB EC

3.2 In the diagram, ADE is a triangle having BC | | ED

and AE | | GF. It is also given that AB : BE = 1 : 3,

AC = 3 units, EF = 6 units, FD = 3 units and

CG = x units.

Calculate, giving reasons: 3.2.1 the length of CD (3)

3.2.2 the value of x (4)

3.2.3 the length of BC (5)

3.2.4 the value of

area ABC

area GFD (5) [23]

F

G

A

E

B

H

D

C

20º

68º

1

1

2

2

2

1

3

3

4

MA

E

B

D

C1 1 2

2

2

1

3

3

4

1

2

3 F

x

(6)

G

F

3

x

63

A

E

B

D

C

A

E

B

D

C


GRADE 12: MEMOS

1.1 . . . the angle subtended by the chord in the alternate

segment.

1.2.1 ˆ

1E = ˆ 1B . . . tan chord theorem

= 68º �

1.2.2 ˆ

3B = ˆ 1E . . . alt. øs ; AE || BC

= 68º �

1.2.3 ˆ

1D = ˆ 3B . . . ext. ø of cyclic quad.

= 68º �

1.2.4 ˆ

2E = ˆ 1D + 20º . . . ext. ø of

= 88º �

1.2.5 ˆC = 180º - ˆ 2E . . . opp. øs of cyclic quad.

= 92º �

2.1 Â = x . . . tan chord theorem

ˆ

2D = x . . . øs opp. equal sides

2.2

ˆ

1M = ˆ ˆ

2A + D . . . ext. ø of

= 2x

â ˆ

2M = 90º - 2x . . . ME AC

& ˆMDE = 90º . . . radius MD tangent CDE

â Ê = 2x . . . sum of øs in MED

â ˆ

1M = Ê â CM is a tangent at M to ?MED � . . .

2.3 ÂDB = 90º . . . ø in semi-?

& ˆ

3M = 90º . . . ME AC

â ˆ

3M = ÂDB â FMBD is a cyclic quad � . . . converse ext. ø of cyclic quad

2.4 Let BC = a ; then MB = 2a

â MD = 2a . . . radii

In MDC: ˆMDC = 90º . . . radius tangent

â DC2 = MC2 - MD2 . . . theorem of Pythagoras

= (3a)2 - (2a)2

= 9a2 - 4a2

= 5a2

= 5BC2 �

2.5 In s DBC and DFM

(1) ˆ

1B = ˆ2F . . . ext ø of c.q. FMBD

(2) ˆ

4D = ˆ 2D . . . both = x

â DBC ||| DFM � . . . equiangular s

2.6 â DM

FM =

DC

BC . . . ||| s

= 5 BC

BC . . . see 2.4

= 5 �

3.1 Construction:

Join DC and EB

and heights h and h′

area of ADE

area of DBE

=

1

2 AD h.

1

2 DB h.

=

AD

DB . . . equal heights

& area of ADE

area of EDC

=

1

2 AE h.

1

2 EC h.

=

AE

EC . . . equal heights

But, area of DBE = area of EDC . . .

â area of ADE

area of DBE

=

area of ADE

area of EDC

â

AD

DB =

AE

EC �

3.2.1 Let AB = p ; then BE = 3p

In AED: CD

3 =

3p

p . . . proportion thm; BC || ED

% 3) â CD = 9 units �

3.2.2 CG = x ; so GD = 9 - x

In DAE: 9 -

+ 3

x

x

= 3

6 . . . prop. thm. ; AE || GF

â 54 - 6x = 3x + 9

â -9x = -45

â x = 5 �

3.2.3 In s ABC and AED

(1) Â is common

(2) ÂBC = Ê . . . corr. øs ; BC || ED

â ABC ||| AED . . . equiangular s

â BC

ED =

AB

AE . . . ||| s

â BC

9 =

p

4p

% 9) â BC = 9

4units �

3.2.4 area of ABC

area of GFD

=

ˆ

ˆ

1

2

1

2

AC BC sin ACB

DG DF sin D

.

.

=

1

2 3. ˆ

9

4

1

2

sin D. .

4 3. .

ˆ

sin D.

. . . corr. øs ; BC || ED

=

9

4

4

= 9

16 �

OR: area of ABC

area of AED

=

1

2 p. 3. ˆ

1

2

sin A.

4p. 12.

ˆ

4

sin A.

= 1

16

â area of ABC = 1

16area of AED . . . �

& area of GFD

area of AED

=

1

2 4. 3. ˆ

sin D.

1

2 12.

3

9. ˆ

3

sin D.

= 1

9

â area of GFD = 1

9area of AED . . . �

� � : â area of ABC

area of GFD

=

1

16 area of AED

1

9 area of AED

= 9

16 �

M A

E

B

D

C 11 2

2

2

1

3

3

4

1

2

3Fx

A

E

B

D

C

h′ h

same base DE &

betw. same || lines,

i.e. same height

Proof :

converse tan chord

theorem

G

F

p

x

6 3

A

E

B

D

C

3p

3

6

PROBLEM-SOLVING: An Active Approach

Be Active . . .

Use all the Clues/Triggers

Recall the Theory systematically

ACT! A

C

T

CAPS Curriculum page 14

THE CAPS CURRICULUM: OVERVIEW OF TOPICS

Paper 2

7. EUCLIDEAN GEOMETRY AND MEASUREMENT

Grade 10

Grade 11

Grade 12

(a) Revise basic results established in earlier

grades. (b) Investigate line segments joining the mid-

points of two sides of a triangle. (c) Properties of special quadrilaterals.

(a) Investigate and prove theorems of the

geometry of circles assuming results from

earlier grades, together with one other result

concerning tangents and radii of circles. (b) Solve circle geometry problems, providing

reasons for statements when required. (c) Prove riders.

(a) Revise earlier (Grade 9) work on the

necessary and sufficient conditions for

polygons to be similar. (b) Prove (accepting results established in earlier

grades): � that a line drawn parallel to one side of a

triangle divides the other two sides

proportionally (and the Mid-point Theorem

as a special case of this theorem); � that equiangular triangles are similar; � that triangles with sides in proportion are

similar; � the Pythagorean Theorem by similar

triangles; and � riders

THE CAPS CURRICULUM TERM BY TERM CONTENT

GRADE 10 - GRADE 12

MATH

EM

ATICS

GR

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10-12

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GRADE 10: TERM 1

Weeks topic Curriculum statement Clarification

3 EuclideanGeometry

1. Revise basic results established in earliergrades regarding lines, angles andtriangles, especially the similarity andcongruence of triangles.

2. Investigate line segments joining the mid-points of two sides of a triangle.

3. Define the following special quadrilaterals:the kite, parallelogram, rectangle, rhombus,square and trapezium. Investigate andmake conjectures about the properties ofthe sides, angles, diagonals and areasof these quadrilaterals. Prove theseconjectures.

Comments:

• Triangles are similar if their corresponding angles are equal, or if the ratios of their sides are

equal: Triangles ABC and DEF are similar if DA ˆˆ = , EB ˆˆ = and FC ˆˆ = . They are also

similar if .

• We could define a parallelogram as a quadrilateral with two pairs of opposite sides parallel.Then we investigate and prove that the opposite sides of the parallelogram are equal, oppositeangles of a parallelogram are equal, and diagonals of a parallelogram bisect each other.

• It must be explained that a single counter example can disprove a Conjecture, but numerousspecific examples supporting a conjecture do not constitute a general proof.

example:

In quadrilateral KITe, KI = Ke and IT = eT. The diagonals intersect at M. Prove that:

1. IM = Me and (R)

2. KT is perpendicular to Ie. (P)

As it is not obvious, first prove that .

MATH

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ATICS

GR

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GRADE 10: TERM 4 Weeks topic Curriculum statement Clarification

2 EuclideanGeometry

Solve problems and prove riders using the properties of parallel lines, triangles and quadrilaterals.

Comment:Use congruency and properties of quads, esp. parallelograms.

example:eFGH is a parallelogram. Prove that MFNH is a parallelogram.

N

M

H E

FG

(C)

E F

GH

M

N

MATH

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GRADE 11: TERM 1no. of weeks topic Curriculum statement Clarification

3 EuclideanGeometry

Accept results established in earlier grades

as axioms and also that a tangent to a circle

is perpendicular to the radius, drawn to the

point of contact.

Then investigate and prove the theorems of

the geometry of circles:

• The line drawn from the centre of a circleperpendicular to a chord bisects the chord;

• The perpendicular bisector of a chordpasses through the centre of the circle;

• The angle subtended by an arc at thecentre of a circle is double the size of theangle subtended by the same arc at thecircle (on the same side of the chord as thecentre);

• Angles subtended by a chord of the circle,on the same side of the chord, are equal;

• The opposite angles of a cyclic quadrilateralare supplementary;

• Two tangents drawn to a circle from thesame point outside the circle are equal inlength;

• The angle between the tangent to a circleand the chord drawn from the point ofcontact is equal to the angle in the alternatesegment.

Use the above theorems and their

converses, where they exist, to solve riders.

Comments:Proofs of theorems can be asked in examinations, but their converses

(wherever they hold) cannot be asked.

example:1. AB and CD are two chords of a circle with centre o. M is on AB and N is on CD such that

OM ⊥ AB and ON ⊥ CD. Also, AB = 50mm, OM = 40mm and ON = 20mm. Determine theradius of the circle and the length of CD. (C)

2. O is the centre of the circle below and xO 2ˆ1 = .

2.1. Determine 2O and M in terms of x . (R)

2.2. Determine 1K and 2K in terms of x . (R)

2.3. Determine MK ˆˆ1 + . What do you notice? (R)

2.4. Write down your observation regarding the measures of 2K and M . (R)

Bonnie

Oval

MATH

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ATICS

GR

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10-12

48C

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GRADE 12 TERM 1no. of Weeks

topic Curriculum statement Clarification

3 Euclidean Geometry

1. Revise earlier work on the necessary andsufficient conditions for polygons to besimilar.

2. Prove (accepting results established inearlier grades):• that a line drawn parallel to one side

of a triangle divides the other twosides proportionally (and the Mid-pointTheorem as a special case of thistheorem) ;

• that equiangular triangles are similar;• that triangles with sides in proportion are

similar; and• the Pythagorean Theorem by similar

triangles.

example:

Consider a right triangle ABC with . Let and .

Let D be on such that . Determine the length of in terms of a and c . (P)

Copyright © The Answer Series

Euclidean Geometry: Theorem Statements & Acceptable Reasons

LINES

The adjacent angles on a straight line are supplementary. øs on a str line

If the adjacent angles are supplementary, the outer arms

of these angles form a straight line. adj øs supp

The adjacent angles in a revolution add up to 360º. øs round a pt OR øs in a rev

Vertically opposite angles are equal. vert opp øs

If AB || CD, then the alternate angles are equal. alt øs ; AB || CD

If AB || CD, then the corresponding angles are equal. corresp øs ; AB || CD

If AB || CD, then the co-interior angles are supplementary. co-int øs ; AB || CD

If the alternate angles between two lines are equal, then

the lines are parallel. alt øs =

If the corresponding angles between two lines are equal,

then the lines are parallel. corresp øs =

If the co-interior angles between two lines are

supplementary, then the lines are parallel. co-int øs supp

TRIANGLES

The interior angles of a triangle are supplementary. ø sum in Δ OR sum of øs OR

int øs in Δ

The exterior angle of a triangle is equal to the sum of the

interior opposite angles. ext øs of Δ

The angles opposite the equal sides in an isosceles

triangle are equal. øs opp equal sides

The sides opposite the equal angles in an isosceles

triangle are equal. sides opp equal øs

In a right-angled triangle, the square of the hypotenuse is

equal to the sum of the squares of the other two sides.

Pythagoras OR

Theorem of Pythagoras

If the square of the longest side in a triangle is equal to

the sum of the squares of the other two sides then the

triangle is right-angled.

Converse Pythagoras OR Converse Theorem of Pythagoras

If three sides of one triangle are respectively equal to

three sides of another triangle, the triangles are

congruent.

SSS

If two sides and an included angle of one triangle are

respectively equal to two sides and an included angle of

another triangle, the triangles are congruent.

SAS OR SøS

If two angles and one side of one triangle are

respectively equal to two angles and the corresponding

side in another triangle, the triangles are congruent.

AAS OR øøS

If in two right angled triangles, the hypotenuse and one side

of one triangle are respectively equal to the hypotenuse

and one side of the other, the triangles are congruent.

RHS OR 90ºHS

The line segment joining the midpoints of two sides of a

triangle is parallel to the third side and equal to half the

length of the third side.

Midpt Theorem

The line drawn from the midpoint of one side of a

triangle, parallel to another side, bisects the third side.

line through midpt || to

2nd side

A line drawn parallel to one side of a triangle divides the

other two sides proportionally.

line || one side of Δ OR

prop theorem; name || lines

If a line divides two sides of a triangle in the same

proportion, then the line is parallel to the third side.

line divides two sides of Δ in

prop

If two triangles are equiangular, then the corresponding

sides are in proportion (and consequently the triangles

are similar).

||| Δs OR equiangular Δs

If the corresponding sides of two triangles are

proportional, then the triangles are equiangular (and

consequently the triangles are similar).

Sides of Δ in prop

If triangles (or parallelograms) are on the same base (or on

bases of equal length) and between the same parallel lines,

then the triangles (or parallelograms) have equal areas.

same base; same height OR

equal bases; equal height

Copyright © The Answer Series xi

QUADRILATERALS

The interior angles of a quadrilateral add up to 360º. sum of øs in quad

The opposite sides of a parallelogram are parallel. opp sides of ||m

If the opposite sides of a quadrilateral are parallel, then

the quadrilateral is a parallelogram.

opp sides of quad are || OR

converse opp sides of ||m

The opposite sides of a parallelogram are equal in length. opp sides of ||m

If the opposite sides of a quadrilateral are equal, then the

quadrilateral is a parallelogram.

opp sides of quad are = OR

converse opp sides of a parm

The opposite angles of a parallelogram are equal. opp øs of ||m

If the opposite angles of a quadrilateral are equal then


opp øs of quad are = OR

converse opp angles of a

parm

The diagonals of a parallelogram bisect each other. diag of ||m

If the diagonals of a quadrilateral bisect each other, then


diags of quad bisect each

other OR

converse diags of a parm

If one pair of opposite sides of a quadrilateral are equal

and parallel, then the quadrilateral is a parallelogram. pair of opp sides = and ||

The diagonals of a parallelogram bisect its area. diag bisect area of ||m

The diagonals of a rhombus bisect at right angles. diags of rhombus

The diagonals of a rhombus bisect the interior angles. diags of rhombus

All four sides of a rhombus are equal in length. sides of rhombus

All four sides of a square are equal in length. sides of square

The diagonals of a rectangle are equal in length. diags of rect

The diagonals of a kite intersect at right-angles. diags of kite

A diagonal of a kite bisects the other diagonal. diag of kite

A diagonal of a kite bisects the opposite angles. diag of kite

CIRCLES

GROUP I

The tangent to a circle is perpendicular

to the radius/diameter of the circle at

the point of contact. tan ⊥ radius

tan ⊥ diameter

If a line is drawn perpendicular to a

radius/diameter at the point where the

radius/diameter meets the circle, then

the line is a tangent to the circle.

line ⊥ radius OR

converse tan ⊥ radius OR

converse tan ⊥ diameter

The line drawn from the centre of a

circle to the midpoint of a chord is

perpendicular to the chord.

line from centre to midpt of chord

The line drawn from the centre of a

circle perpendicular to a chord bisects

the chord.

line from centre ⊥ to chord

The perpendicular bisector of a

chord passes through the centre of

the circle.

perp bisector of chord

The angle subtended by an arc at the

centre of a circle is double the size of

the angle subtended by the same arc

at the circle (on the same side of the

chord as the centre)

øat centre

= 2 % ø at circumference

The angle subtended by the diameter

at the circumference of the circle

is 90º.

øs in semi circle OR

diameter subtends right angle

OR ø in ½ ?

If the angle subtended by a chord at

the circumference of the circle is 90°,

then the chord is a diameter.

chord subtends 90º OR

converse øs in semi circle

O

O

2x

x

O

O

O

O

O

O

Copyright © The Answer Series xii

GROUP II

Angles subtended by a chord of the

circle, on the same side of the

chord, are equal øs in the same seg

If a line segment joining two points

subtends equal angles at two

points on the same side of the line

segment, then the four points are

concyclic. (This can be used to prove that the

four points are concyclic).

line subtends equal øs

OR

converse øs in the same seg

Equal chords subtend equal angles

at the circumference of the circle. equal chords; equal øs


at the centre of the circle. equal chords; equal øs

Equal chords in equal circles subtend

equal angles at the circumference of

the circles.

equal circles; equal chords;

equal øs

Equal chords in equal circles

subtend equal angles at the centre

of the circles. (A and B indicate the centres of

the circles)

equal circles; equal chords;

equal øs

GROUP III

The opposite angles of a cyclic

quadrilateral are supplementary

(i.e. x and y are supplementary)

opp øs of cyclic quad

If the opposite angles of a

quadrilateral are supplementary

then the quadrilateral is cyclic.

opp øs quad sup OR

converse opp øs of cyclic quad

The exterior angle of a cyclic

quadrilateral is equal to the

interior opposite angle.

ext ø of cyclic quad

If the exterior angle of a quadrilateral

is equal to the interior opposite

angle of the quadrilateral, then the

quadrilateral is cyclic.

ext ø = int opp ø

OR

converse ext ø of cyclic quad

GROUP IV

Two tangents drawn to a circle

from the same point outside the

circle are equal in length (AB = AC)

Tans from common pt

OR

Tans from same pt

The angle between the tangent to

a circle and the chord drawn from

the point of contact is equal to the

angle in the alternate segment.

tan chord theorem

If a line is drawn through the end-

point of a chord, making with the

chord an angle equal to an angle

in the alternate segment, then the

line is a tangent to the circle.

(If x = b or if y = a then the

line is a tangent to the circle)

converse tan chord theorem

OR

ø between line and chord

x

xy

y

x x

xx

O x x

xx

A

x B

x

x

y

x

180º – x

x

x

x

x

b x a

y

B

A

C

x x

y

y


IMPORTANT ADVICE FOR MASTERING MATHS

Don't focus on what you haven't done in the past.

Put that behind you and start today! Give it your all – it is well worth it!

� TIMETABLE / PLANNING

■ Draw up a timetable of study times.

■ Revise your schedule from time to time to ensure optimum focus and

awareness of time.

Motivation will not be a problem once you've done this,

because you will see that you need to use every minute!

� ROUTINE

Routine is really important. Start early in the morning, at the same time every

day, and don't work beyond 11 at night. Arrange some 1 hour and some 2 hour sessions on

particular subjects. Schedule more difficult pieces of work for early in the day

and easier bits for later when you're tired. Reward yourself with an early night now and again! Allow some time for physical exercise – at least ½ hour a

day. Any sport or walking, jogging (or skipping when it rains)

will improve your concentration.

� 'NO-NO'S'

Limit the time you spend

■ on your phone

■ at the television

■ on Facebook and any other social networks

■ in the sun

All these activities break down your commitment,

focus and energy.

� WORK FOCUS

Don't worry about marks! Just focus on the work and the marks will take care

of themselves.

Worrying is tiring and time-wasting and gets in the way of your progress!

Your marks will gradually improve if you work consistently.

� YOUR APPROACH

The most important thing of all is to remain positive. Some times will be

tough, some exams WILL BE TOUGH, but in the end, your results will reflect

all the effort that you have put in.

■ Despair can destroy your Mathematics

Mathematics should be taken on as a continual challenge (or not at all!).

Teach your ego to suffer the 'knocks' which it may receive – like a poor

test result. Instead of being negative about your mistakes (e.g. 'I'll never

be able to do these sums'), learn from them. As you address each one,

they will help you to understand the work and do better next time!

■ Work with a friend occasionally.

Discussing Mathematics makes it alive and enjoyable.

� A GREAT GENERAL STUDY TIP

■ Don't just read through work!

■ Study a section and then, on your own, write down all you can remember.

Knowing that you're going to do this makes you study in a logical, alert

way. ■ You're then only left to learn the few things which you left out. ■ This applies to all subjects. In this text, read the explanations very

carefully and actively, trying all worked examples yourself first as you

master each topic. ■ A subject like Maths also requires you to practise and apply the concepts

regularly.


� ABOUT THE MATHS

■ Try each problem on your own first – no matter how inadequately –

before consulting the answer. It is only by encountering the difficulties

which you personally have that you will be able, firstly, to pin-point them,

and then, secondly, to understand and rectify them and then make sure

you don't make them again!!

■ Learn to keep asking yourself 'why'?

It is when you learn to REASON that you really start enjoying Maths and,

quite coincidentally, start doing well at it!!

Answers are by no means the most important thing in Mathematics.

When you've done a problem, don't be satisfied only to check the answer.

Check also on your layout and reasoning (logic). Systematic, to-the-point,

logical and neat presentation is very important.

■ Revise earlier work as often as possible. Set up a revision plan with at

least one session a week for this purpose. Familiarity is the key to success

in Maths!

EXAM PREPARATION

The best way to prepare for your exam is to start early – in fact, on the

first day of the year!!! Working past papers is excellent preparation for any

exam – and The Answer Series provides these – but, only when you're ready.

Focus on WORKING ON ONE TOPIC AT A TIME first. It is the most effective

way to improve, particularly as you will build up your confidence this way.

The Answer Series provides thorough topic treatment for all subjects,

enabling you to cover all aspects of each topic, from the basics to the top

level questions. Thereafter, working past papers is a worthwhile and

rewarding exercise.

THE EXAMS

Finally, for the exams themselves, make sure you have all you need (e.g. your

calculator, ruler, etc.) and don't allow yourself to be upset by panicking friends.

Plan your time in the exam well – allowing some time to check at the end.

Whatever you do, don't allow yourself to get stuck on any difficult issues in the

exam. Move on, and rather come back to problem questions if you have time

left. If you're finding an exam difficult, just continue to do your absolute best

right until the end! Partial answers can earn marks.

We wish you the best of luck in your studies and hope that this book

will be the key to your success – enjoy it!!

The Answer Series Maths Team

PLEASE NOTE:

These Geometry materials (Booklets 1 to 4) were created

and produced by The Answer Series Educational Publishers

(Pty) (Ltd) to support the teaching and learning of Geometry

in high schools in South Africa.





GRADES 8 - 12ALL MAJOR SUBJECTS IN

ENGLISH & AFRIKAANS

WWW.THEANSWER.CO.ZA

1

Philosopher, Immanuel Kant (18th century philosopher)

Theory without practice

is empty

Practice without theory

is blind

2

CONTENT FRAMEWORK

LINES

TRIANGLES

QUADRILATERALS

CIRCLES (Gr 11)

(Gr 8 10)

Gr 12?

3

Gr 12: Theorem of Pythagoras (Gr 8)

Similar Δs (Gr 9)

Midpoint Theorem (Gr 10)

& The Proportion Theorem

Ratio Proportion Area

4

LINES

2 Situations

Converse Statements

Logic !

NB:

Vocabulary first,

then facts

2 3 1

4

1 2

3 4

5 6

7 8

1

2

5

TRIANGLES

Sum of Interior øs Exterior ø of Δ

Isosceles Δ Equilateral Δ Right-ød Δ

– Theorem of Pythagoras

Area of a Δ and related facts

Similar Δs Congruent Δs

Midpoint Theorem

NB:

Vocabulary first,

then facts

These involve

converse theorems

�

�

�

�

�

Copyright © The Answer

QUADRILATERALS - definitions, areas & properties

'Any' Quadrilateral

A Trapezium

A Parallelogram

The Square

A Rhombus

A Kite

DEFINITION:

Quadrilateral with 1 pair of opposite sides ||

DEFINITION:

Quadrilateral with 2 pairs opposite sides ||

DEFINITION:

A ||m with one pair of

adjacent sides equal

the 'ultimate' quadrilateral !

Sum of the øs of

any quadrilateral = 360º Area = s2

Area = Δ 1 + Δ 2

= 1

2ah +

1

2bh

= 1

2(a + b) .h

'Half the sum of the || sides

% the distance between them.'

Properties :

It's all been said 'before' !

Since a square is a rectangle,

a rhombus, a parallelogram,

a kite, . . . ALL the properties

of these quadrilaterals apply.

Area

= 1

2product of diagonals (as for a kite)

or

= base % height (as for a parallelogram)

Given diagonals a and b

Area = 2Δs = 2

1 ab

2 2

⎛ ⎞⎜ ⎟⎝ ⎠

. = ab

2

'Half the product of the diagonals'

THE DIAGONALS

• bisect one another PERPENDICULARLY

• bisect the angles of the rhombus

• bisect the area of the rhombus

a b

c

d e

f

h

b

a

1

2

b

a

2

a

2

Sum of the interior angles

= (a + b + c) + (d + e + f)

= 2 % 180º . . . ( 2 Δs )

= 360º

Properties :

2 pairs opposite sides equal

2 pairs opposite angles equal

& DIAGONALS BISECT ONE ANOTHER

Area = base % height

||m ABCD = ABCQ + ΔQCD

rect. PBCQ = ABCQ + ΔPBA

where ΔQCD ≡ ΔPBA . . . RHS/90ºHS

â ||m ABCD = rect. PBCQ (in area)

= BC % QC

P A Q

D

B C

All you need to know

DEFINITION:

Quadrilateral with 2 pairs of adjacent sides equal

A Rectangle

DEFINITION:

A ||m with one right ø

Area = ℓ % b

DIAGONALS are EQUAL

angles

diagonals

sides The arrows indicate

various ‘pathways’

from ‘any’

quadrilateral to the

square (the ‘ultimate

quadrilateral’). These

pathways, which combine

logic and fact, are

essential to use when

proving specific types

of quadrilaterals.

See how the properties

accumulate as we

move from left to right,

i.e. the first quad. has

no special properties

and each successive

quadrilateral has all

preceding properties.

Quadrilaterals play a prominent role in both Euclidean & Analytical Geometry right through

to Grade 122!

Note :

y y

y y x

x

x

x

2x + 2y = 180º . . .

� x + y = 90º

THE DIAGONALS

• cut perpendicularly

• ONE DIAGONAL bisects the other diagonal,

the opposite angles and the area of the kite ø

s of Δ or

co-int. øs ; || lines


2x

x x

x

2x

x + y = 180º

y

x

There are ' 2 ways to prove that a line is a tangent to a ? '.

SUMMARY OF CIRCLE GEOMETRY THEOREMS

TThhee

''CCeennttrree''

ggrroouupp

TThhee

''TTaannggeenntt''

ggrroouupp

TThhee

''CCyycclliicc QQuuaadd..''

ggrroouupp

TThhee

''NNoo CCeennttrree''

ggrroouupp

Equal

radii !

There are ' 3 ways to prove that a quad. is

a cyclic quad '.

III

I

IV

II

Equal

chords!

Equal

tangents!

Lines, Angles, Δs & Quadrilaterals(Grade 8 to 10 Revision)


TAS FET EUCLIDEAN GEOMETRY COURSEBOOKLET 2

Lines, Angles, Δs & Quadrilaterals


WWW.THEANSWER.CO.ZA

CONTENTS of Booklet 2 Lines & Angles

Vocabulary and Facts

Triangles

Vocabulary, Facts and Proofs Area

Quadrilaterals

Properties Definitions Theorems Area

Grade 10: Midpoint Theorem

Statement and Converse Riders

EXERCISES & FULL SOLUTIONS on all 4 sections


The Language (Vocabulary)

� PARALLEL LINES � PERPENDICULAR LINES

AB | | CD AB ⊥ CD

� SINGLE ANGLES

� PAIRS OF ANGLES

• Complementary øs add up to 90º

e.g. 40º and 50º; x and 90º – x

• Supplementary øs add up to 180º

e.g. 135º and 45º; x and 180º – x

• Adjacent øs have a common vertex

and a common arm and lie on

opposite sides of the common arm.

e.g. A pair of adjacent A pair of adjacent

complementary øs : supplementary øs :

WHEN 2 LINES INTERSECT, WE HAVE . . .

( adjacent øs & ( vertically opposite øs,

1 and 4ˆ ˆ ; 1 and 3ˆ ˆ

4 and 3ˆˆ , etc. or 2 and 4ˆ ˆ

WHEN 2 LINES ARE CUT BY A TRANSVERSAL . . .

2 'families' of 4 angles are formed

� PAIRS OF ANGLES , one from each family:

• corresponding øs : 1 & 5 ; 2 & 6 ; 3 & 7 ; 4 & 8

• alternate øs : 3 & 5 and 4 & 6

• co-interior øs : 4 & 5 and 3 & 6

( Corresponding means their positions correspond.

( Alternate means:

øs lie on opposite sides of the transversal;

whereas, 'co-' means: 'on the same side of the transversal'.

e.g. corresponding alternate co-interior angles angles angles

The ANGLE is the amount of rotation about the vertex.

acute angles

obtuse angles

reflex angles

0º 90º 180º 270º 360º

a right angle a straight angle a revolution

A

B

C

D A B

C D

arms vertex

NOTE: The plural of vertex is vertices!

40º 50º

x 90º – x

135º 45º

LINES, ANGLES & TRIANGLES

A

C B

D

D

A

C B

90º

40º

140º

50º

40º

!

Know the meanings

of all the WORDS.

1 2 3 4

5 6 7 8

1

23

4

x

180º – x

7

1 2

4 3

5 6

8 common arm

common

vertex

LINES: 2 SITUATIONS

the

transversal

1 2

4 3

5 6

8 7

2 Copyright © The Answer Series

Classification according to . . .

SIDES ANGLES

� Scalene Δ � Acute ød Δ

(all 3 sides different in length) (all 3 øs are acute)

� Isosceles Δ � Right ød Δ

(2 sides equal in length) (one ø = 90º)

� Equilateral Δ � Obtuse ød Δ

(all 3 sides equal in length) (one ø is obtuse) In an isosceles triangle:

We can classify Δs according to sides and øs simultaneously:

Examples

This is a/an . . . . -angled Δ This is a/an . . . . -angled Δ This is a/an . . . . -angled Δ

The Facts

LINES

INTERSECTING LINES

When two lines intersect, any pair of adjacent angles is

supplementary.

e.g. 4ˆ + 1ˆ = 180º & 4ˆ + 3ˆ = 180º

Vertically opposite angles are equal.

e.g. 1ˆ = 3ˆ and 2ˆ = 4ˆ

PARALLEL LINES

When ANY 2 lines are cut by a transversal, 2 'families' of four øs are formed,

and there are:

corresponding øs ;

alternate øs ;

& co-interior øs

. . . whether the lines are parallel or not !

If 2 PARALLEL lines are cut by a transversal,

� corresponding angles are equal ;

� alternate angles are equal ; and

� co-interior angles are supplementary.

1 2 3 4

5 6

7 8

1 2

3 4

2 intersecting lines

form 4 angles

the vertical angle

the base angles

BASE

Fact 2

Fact 1

Fact 3

The sizes of the other angles?

Special case:

Perpendicular lines

TRIANGLES

Answers

An isosceles right-ød Δ

An isosceles acute-ød Δ

A scalene obtuse-ød Δ

Interior and Exterior angles

Interior angles: & Exterior angles:

x

y

z a

c

b

d

f

e

or

An exterior angle

is formed between

a side of the triangle

and the produced

(extension) of

an adjacent side of

the triangle. x

x is not an exterior ø

. . . the side is not 'produced'.

1 2

3 4

5 6

7 8


The CONVERSE STATEMENT of FACT 3 says

If a pair of corresponding angles is equal

OR

If a pair of alternate angles is equal

OR

If a pair of co-interior angles is supplementary

then:

line AB is PARALLEL to line CD

. . . whether it looks like it or not!

FACT 1 says:

If ABC is a straight line, then 1ˆ + 2ˆ = 180º.

Example

If, in the sketch alongside, x = 40º and y = 130º,

is PQR a straight line?

Now, refer to FACT 3 on the previous page . . .

The Facts, continued

TRIANGLES

FACT 1: The Sum of the interior angles of a triangle . . .

The sum of the (interior) angles

of a triangle is 180º.

FACT 2: The Exterior angle of a triangle . . .

The exterior angle of a triangle equals

the sum of the interior opposite angles.

FACT 3: An Isosceles triangle . . .

In an isosceles triangle,

the base angles are equal.

Example

If in isosceles ΔABC,

the vertical angle (A)ˆ is a right ø,

determine the size of the base øs.

If AB = AC,

then: 1ˆ = 2ˆ

y

x

P

Q

R

CONVERSE STATEMENTS

Aˆ + Bˆ + Cˆ = 180º

A

B C

1 2

A

B C

1ˆ = 2ˆ + 3ˆ2

3 1

The CONVERSE STATEMENT of FACT 1 says

If 1ˆ + 2ˆ = 180º, then: ABC is a straight line.

A

B

C D

So, conversely:

If 1ˆ = 2ˆ ,

then: AB = AC

The CONVERSE states:

If 2 angles of a triangle are equal,

then: the sides opposite them are equal.

x

A

B Cx

1

2

A

B

C

Answer: x + y = 40º + 130º = 170º ≠ 180º

â No, PQR is not a straight line . . . Whether it looks

like it or not!

Answer

2x = 90º

∴ x = 45º



If the square on one side

of a triangle equals the

sum of the squares on the

other two sides (in area),

then the triangle is

right-angled.

FACT 4: An Equilateral triangle . . .

The angles of an equilateral triangle

all equal 60º.

FACT 5: THE THEOREM OF PYTHAGORAS

THE THEOREM OF PYTHAGORAS states:

The square on the hypotenuse of a right-angled triangle

equals the sum of the squares on the other two sides (in area).

Note:

• Only one angle can be 90º (a right angle)

• The side opposite the right-angle is called the hypotenuse. FACT 6: The Area of a triangle

The area of a triangle = base height

2

%

FACT 7: Similar triangles

Definition of Similarity

Two figures are SIMILAR if :

they are equiangular, and

their corresponding sides are in proportion.

e.g.

In the case of TRIANGLES (only):

If triangles are equiangular,

then: their corresponding sides are in proportion

and therefore: they are similar.

i.e. If Aˆ = Dˆ , Bˆ = Eˆ and Cˆ = F,ˆ

then: AB

DE =

BC

EF =

AC

DF,

and therefore: ΔABC ||| ΔDEF.

60º

60º 60º

D

E F

A

B C

3x = 180º

∴ x = 60º


If the sides of two triangles are in proportion,

then: these triangles will also be equiangular

and therefore: the triangles are similar.

i.e. If ABDE

= BC

EF =

AC

DF,

then: Aˆ = Dˆ , Bˆ = Eˆ and Cˆ = Fˆ ,

and therefore: ΔABC ||| ΔDEF

If ˆC = 90º,

then:

c2 = a2 + b2

Conversely:

If c2 = a2 + b2,

then:

ˆC = 90º

A

CB a

b c

h

b

The symbol: |||

triangles quadrilaterals pentagons

. . .

2 conditions


FACT 8: Congruent triangles

Two triangles are congruent if they have

• 3 sides the same length . . . SSS

• 2 sides & an included angle equal . . . SøS

• a right angle, the hypotenuse & a side equal . . . RHS or SS90º

• 2 angles and any side equal . . . øøS

If we can prove that ΔABC ΔDEF, then we can conclude that

all the sides and angles are equal.

Congruency can be understood best by constructing triangles,

even casually imagining the construction!

The possibilities are:

SSS – only one size (& shape) of triangle could

be constructed.

SSø – the case where the angle is NOT INCLUDED

between the sides.

This is the ambiguous case because

there are 2 possible ∆s

which we could draw

RHS – the angle is not included,

but it is a right angle,

so only 1 option is possible:

SøS – the angle is included.

Only 1 option is possible.

øøS – given 2 angles, we actually have all 3 angles since the sum of

the angles must be 180º.

The side restricts the size of the triangle.

So, only 1 option is possible.

However, when comparing triangles,

the given equal sides must correspond

in relation to the angles.

øøø – A side is

required!

Any number of possible options.

Congruent Δs have the same SHAPE, and the same SIZE.

Similar Δs have the same SHAPE, but not necessarily the same SIZE.

or

or

obtuse ød ∆ acute ø

d ∆

A

B C

D

F E

or

The symbol:

Conditions of Congruency

A triangle has 6 parts, 3 sides and 3 angles, which can be measured.

However, we use 3 measurements at a time to construct a triangle.


( If 2 triangles are equal in every respect (i.e. all 3 sides and

all 3 angles), we say they are congruent.

We write: ABC PQR.

( Enlarging or reducing a triangle, as on a copier, will produce a triangle

similar to the original one.

i.e. It will have the same shape (all angles will be the same size as before)

but the respective sides will not be the same length.

The sides will, however, be proportional.

We write: ABC ||| PQR.

Comparing Triangles – Congruence vs Similarity

Proofs of Triangle Facts 1 and 2

Triangle Fact 1

Why is the sum of the interior angles of a triangle 180º?

See ΔABC:

Can we prove that Bˆ + Aˆ + Cˆ = 180º ?

A

B C

Draw line DAE through A, parallel to BC.

We know: 1ˆ + 2ˆ + 3ˆ = 180º . . . adjacent øs on a straight line, DAE

But ˆ1 = alternate ˆB & ˆ3 = alternate ˆC . . . DAE || BC

â ˆB + ˆ2 + ˆC = 180º

DA

B C

E 1

23

Triangle Fact 2

Why is the exterior angle of a triangle equal to

the sum of the two interior opposite angles?

x + y = 180º . . . see LINES fact 1 (Page 2)

but x + ˆ ˆ(A + B) = 180º . . . see TRIANGLES fact 1 (Page 3)

â y = ˆ ˆA + B Logical !

A

B C D

x y


QUADRILATERALS

The Facts

� Properties of Quadrilaterals

• Recall all the quadrilaterals ... (kite, trapezium, parallelogram, rectangle, rhombus, square).

• What properties do they have?

� Equal sides?

� Parallel sides?

� Angles?

Which are equal? Which are supplementary? Which are right angles?

� Diagonals?

Investigating quadrilaterals, using diagonals:

fig. 1: Use a diagonal to determine the sum of

the interior angles of a quadrilateral. fig. 2: Use a diagonal to find the area of

a trapezium. fig. 3 - 6: Which of these quadrilaterals have their areas bisected by

the diagonal? fig. 3 - 6: Draw in the second diagonal. For each figure, establish whether the

diagonals are: � equal � bisect each other

� intersect at right angles � bisect the angles of the quadrilateral fig. 6: Find the area of a kite in terms of its diagonals.

Could this formula apply to a rhombus? A square?

� Defining Quadrilaterals

� A trapezium

� A parallelogram

We have observed the properties of a parallelogram:

� both pairs of opposite sides parallel

� both pairs of opposite sides equal

� both pairs of opposite angles equal

� diagonals which bisect one another.

Observe the progression of quadrilaterals below as we discuss further definitions:

• Which property does a parallelogram need to become a rectangle?

• Which property does a parallelogram need to become a rhombus?

• Which property does a rectangle need to become a square?

• Which property does a rhombus need to become a square?

square parallelogram rhombus kite rectangle

A diagonal of a quadrilateral is a line joining

opposite vertices.

rectangle

rhombus

trapezium parallelogram square

'any' quadrilateral

How would you find the sum

of the interior angles of

a pentagon? A hexagon?

We will, however, define the

parallelogram in terms of its parallel lines.

'any' quadrilateral kite

rhombus square

Definition: A parallelogram is a quadrilateral with TWO PAIRS OF OPPOSITE SIDES parallel.

or

1 3

5

6a 6b

2

4a

4b

trapezium parallelogram square 'any'

quadrilateral

rectangle

rhombus

Definition: A trapezium is a quadrilateral with ONE PAIR OF OPPOSITE SIDES parallel.


� A rectangle

� A rhombus

� A square

All these definitions are extremely important to know when you're doing sums. e.g. If asked to prove that a particular quadrilateral is a rectangle, and you already

know it is a parallelogram, all you need to show is that one angle is a right angle. Observe the following progression of quadrilaterals before the next definitions:

� A kite

Is a rhombus a kite? A rhombus is a kite with .....................................................................................

A square is a kite with ........................................................................................

See the summary of quadrilaterals on the next page:

'Pathways of definitions, areas and properties'

Observing the progression of quadrilaterals – along routes 1 & 2 – is essential for

understanding definitions, properties (especially diagonals) and area formulae.

None of these facts and proofs should need to be memorised.

� Proving conjectures

e.g. 1 Conjecture: The diagonals of a rhombus bisect the angles of the rhombus.

Proof: x1 = x2 . . . øs opp = sides

= x3 . . . alt øs; || lines

= x4 . . . øs opp = sides

e.g. 2 Conjecture: The diagonals of a rhombus intersect at right angles.

(The proof is on the SUMMARY on the next page)

Note:

The further back you go on the route,

the more the definition requires!

Note: The curriculum says:

• Define the quadrilaterals.

• Investigate and make conjectures about the properties of the

sides, angles, diagonals and areas of these quadrilaterals. • Prove these conjectures.

x2

x1

x3

x4

'any' quadrilateral kite

rhombus square

Definition: A square is a rhombus with . . . . . . . . . . . . . .

A square is a rectangle with . . . . . . . . . . . . . .

A square is a parallelogram with . . . . . . . . . . . . . .

A square is a quadrilateral with . . . . . . . . . . . . . .

Definition: A rectangle is a parallelogram with 1 right angle

OR: A rectangle is a quadrilateral with . .?. . right angles?

Definition: A rhombus is a parallelogram with a pair

of adjacent sides equal OR: A rhombus is a quadrilateral with . .?. . equal sides?

Definition: A kite is a quadrilateral with 2 pairs of adjacent sides equal.

Definitions tell us what the minimum (least) is that one needs!

Copyright © The Answer

QUADRILATERALS - definitions, areas & properties

'Any' Quadrilateral

A Trapezium

A Parallelogram

The Square

A Rhombus

A Kite

DEFINITION:

Quadrilateral with 1 pair of opposite sides ||

DEFINITION:

Quadrilateral with 2 pairs opposite sides ||

DEFINITION:

A ||m with one pair of

adjacent sides equal

the 'ultimate' quadrilateral !

Sum of the øs of

any quadrilateral = 360º Area = s2

Area = Δ 1 + Δ 2

= 1

2ah +

1

2bh

= 1

2(a + b) .h



Properties :

It's all been said 'before' !





Area

= 1

2product of diagonals (as for a kite)

or



Area = 2Δs = 2

1 ab

2 2

⎛ ⎞⎜ ⎟⎝ ⎠

. = ab

2


THE DIAGONALS

• bisect one another PERPENDICULARLY

• bisect the angles of the rhombus

• bisect the area of the rhombus

a b

c

d e

f

h

b

a

1

2

b

a

2

a

2

Sum of the interior angles

= (a + b + c) + (d + e + f)

= 2 % 180º . . . ( 2 Δs )

= 360º

Properties :

2 pairs opposite sides equal

2 pairs opposite angles equal

& DIAGONALS BISECT ONE ANOTHER


||m ABCD = ABCQ + ΔQCD

rect. PBCQ = ABCQ + ΔPBA

where ΔQCD ≡ ΔPBA . . . RHS/90ºHS


= BC % QC

P A Q

D

B C

All you need to know

DEFINITION:

Quadrilateral with 2 pairs of adjacent sides equal

A Rectangle

DEFINITION:

A ||m with one right ø

Area = ℓ % b

DIAGONALS are EQUAL

angles

diagonals

sides The arrows indicate

various ‘pathways’

from ‘any’

quadrilateral to the

square (the ‘ultimate

quadrilateral’). These

pathways, which combine

logic and fact, are

essential to use when

proving specific types

of quadrilaterals.

See how the properties

accumulate as we

move from left to right,

i.e. the first quad. has

no special properties

and each successive

quadrilateral has all

preceding properties.

Quadrilaterals play a prominent role in both Euclidean & Analytical Geometry right through

to Grade 122!

Note :

y y

y y x

x

x

x

2x + 2y = 180º . . .

� x + y = 90º

THE DIAGONALS

• cut perpendicularly

• ONE DIAGONAL bisects the other diagonal,

the opposite angles and the area of the kite ø

s of Δ or

co-int. øs ; || lines

2

Defining Quadrilaterals

� A trapezium

� A parallelogram

We have observed the properties of a parallelogram:

� both pairs of opposite sides parallel

� both pairs of opposite sides equal

� both pairs of opposite angles equal

� diagonals which bisect one another.

We will, however, define the parallelogram in terms of

its parallel lines.

Definition: A parallelogram is a quadrilateral with

TWO PAIRS OF OPPOSITE SIDES parallel.

Definition: A trapezium is a quadrilateral with

ONE PAIR OF OPPOSITE SIDES parallel.

These slides on Quadrilaterals are from

The Answer Series Gr 12 Video series

on Analytical Geometry.

3

Observe the progression below as we discuss further definitions . . .

Which property does . . .

1. a parallelogram need to become a rectangle?

2. a parallelogram need to become a rhombus?

3. a rectangle need to become a square?

4. a rhombus need to become a square?

5. And, which property(s) does a parallelogram need to become a square?

'any' quadrilateral trapezium

rectangle

square

rhombus

parallelogram

4

QUADRILATERALS

Consider

� Sides: parallel? equal?

� Angles: equal? supplementary? right ?

� Diagonals . . . . ?

The arrows indicate various

‘pathways’ from ‘any’

quadrilateral to the square

(the ‘ultimate quadrilateral’)

A Trapezium A Parallelogram

The Square

A Rhombus

A Kite

A Rectangle

h

b

a

1

2

b

a

2 a

2

'Any' Quadrilateral

ab

c de

f

The 'ultimate'quadrilateral

5

Investigating diagonals . . .

'Any quadrilateral'

How do we find the sum of the interior angles of a quadrilateral? And a pentagon?

And a hexagon?

Pause now while you continue the pattern

No. of Sides No. of Diagonals No. of Triangles Sum of the interior s

A quadrilateral 4 1 2 2 % 180º = 360º

A pentagon

A hexagon

A polygon with n sides

The sum of the interior angles = (a + b + c) + (d + e + f)

= 2 % 180º . . . (2 s)

= 360º

ab

c

de

f

6

A Trapezium

Can you derive a formula for the area of a trapezium?

The area of a trapezium:

Half the sum of the || sides % the distance between them.

The Area = 1 + 2

= 1

2ah +

1

2bh

= 1

2(a + b).h

h

b

a

1

2

7

A Kite

Can you derive a formula for the area of a kite?

Could this formula apply to a rhombus? And to a square?

Given diagonals a and b . . .

Area = 2s = 2

1 ab

2 2

. = ab

2 . . .

the product of the diagonals

2

The area of a kite: 'Half the product of the diagonals'

A Kite

b

a

2 a

2

8

Now, draw in a second diagonal . . .

Consider each quadrilateral

Determine in which quadrilaterals the diagonals:

1. bisect each other?

2. intersect at right angles?

3. bisect the angles of the quadrilateral?

4. are equal?

A Parallelogram

A Rectangle

A Rhombus

A Square

A Kite

9

The diagonals . . . .

1. bisect one another?

� Rectangle

� Parallelogram � Square

� Rhombus

2. intersect at right angles?

� Kite � Rhombus � Square

3. bisect the s of the quadrilateral?

� Rhombus � Square

4. equal?

� Rectangle � Square

A Rhombus

The Square A Parallelogram

A Rectangle

A Kite

10

A SUMMARY: DIAGONALS

Kite

� cut perpendicularly, and

� the long diagonal bisects: � the short diagonal � the opposite angles & � the area of the kite

Parallelogram Rectangle

� bisect one another � are equal

Rhombus

� bisect one another perpendicularly . . .

� bisect the angles of the rhombus

� bisect the area of the rhombus

Square

� Since a square is a rectangle, a rhombus, a parallelogram, a kite . . .

. . . ALL the properties of these quadrilaterals apply.

2x + 2y = 180º . . . s

of , or

� x + y = 90º co-int. s

suppl.

The s at the point of intersection

of the diagonals are right angles.

y y

y y x

x

x

x

11


||m ABCD = ABCQ + QCD

rect. PBCQ = ABCQ + PBA

where QCD ≡ PBA . . . SS90º


= BC % QC

PA Q

D

B C

SUMMARY: AREAS

A Trapezium

A Parallelogram

The Square

A Rhombus

Area = s2

Area = 1 + 2

= 1

2ah +

1

2bh

= 1

2(a + b).h







A Rectangle

Area = ℓ % b


Area = 2s = 2

1 ab

2 2

. = ab

2


Area of a rhombus

= 12

product of diagonals (as for a kite)

or


h

b

a

1

2

A Kite

b

a

2 a

2


� AN ASSIGNMENT �

TASK A: Theorems 1 → 3

Prove each of these properties yourself,

STARTING WITH THE DEFINITION as the 'given'.

TASK B: Theorems 4 → 7

Prove these four converse theorems,

WORKING TOWARDS THE DEFINITION,

i.e. you need to prove, given any one of these situations, that the quadrilateral

would have 2 pairs of opposite sides parallel, i.e. that, by definition, the

quadrilateral is a parallelogram.

Hint

Use your FACTS on II lines

and congruent triangles.

� Theorems and Proofs

The following section deals with the properties of a parallelogram. We firstly prove

all the properties. Secondly, we prove that a quadrilateral with any of these

properties has to be a parallelogram.

Geometry is an exercise in LOGIC. Initially, we observe, we measure, we

record . . . But, finally . . . We decide on how to define something and

then we prove various properties logically, using the definition.

Beyond the DEFINITION of a parallelogram, we noticed other facts/properties

regarding the lines, angles and diagonals of a parallelogram. The statement and

proofs of these properties make up our first three THEOREMS!

The PROPERTIES of a parallelogram

Theorem 1: The opposite angles of a parallelogram are equal.

Theorem 2: The opposite sides of a parallelogram are equal.

Theorem 3: The diagonals of a parallelogram bisect one another.

The CONVERSE theorems

Given a property, prove the quadrilateral is a parallelogram,

i.e. prove both pairs of opposite sides are parallel. There are four converse statements, each claiming that IF a quadrilateral

has a particular property, it must be a parallelogram.

Theorem 4: If a QUADRILATERAL has 2 pairs of opposite angles equal,

then the quadrilateral is a parallelogram.

Theorem 5: If a QUADRILATERAL has 2 pairs of opposite sides equal,


Theorem 6: If a QUADRILATERAL has 1 pair of opposite sides equal and

parallel, then the quadrilateral is a parallelogram.

Theorem 7: If a QUADRILATERAL has diagonals which bisect one another,


In these cases, we work towards the definition !

THE DEFINITION OF A PARALLELOGRAM

A parallelogram is a quadrilateral with

2 PAIRS OF OPPOSITE SIDES PARALLEL.

All the properties are to be deduced from the definition!


� The Theorem Proofs THE PROOFS OF THE PROPERTIES

� Theorem 1: The opposite angles of a ||m are equal.

Given: ||m ABCD

i.e. AB || DC and AD || BC

RTP: Â = ˆC and ˆB = ˆD

Proof: Â + ˆB = 180º . . . co-interior øs; AD || BC

But, Â + ˆD = 180º . . . co-interior øs; AB || DC

â ˆB = ˆD

Similarly, Â = ˆC

� Theorem 2: The opposite sides of a ||m are equal.

Given: ||m ABCD

i.e. AB || DC and AD || BC RTP: AB = CD and AD = BC Construction: Draw diagonal AC . . . Proof: In Δs ABC and ADC

1) ˆ1 = ˆ2 . . . alternate øs; AB || DC

2) ˆ3 = ˆ4 . . . alternate øs; AD || BC

3) AC is common

â ΔABC ≡ ΔCDA . . . øøS

â AB = CD and AD = BC

� Theorem 3: The diagonals of a parallelogram bisect

one another. Given: ||m ABCD with diagonals AC and BD intersecting at O.

RTP: AO = OC and BO = OD

Proof: In Δs AOB and DOC

1) ˆ ˆ1 = 2 . . . alt øs; AB || DC

2) ˆ ˆ3 = 4 . . . vert opp øs

3) AB = DC . . . opposite sides of ||m – see theorem 2 above

â ΔAOB ≡ ΔCOD . . . øøS

â AO = OC and BO = OD THE CONVERSE PROOFS

� Theorem 4: If a QUADRILATERAL has 2 pairs of opposite

angles equal, then the quadrilateral is a ||m

Given: Quadrilateral ABCD with Â = ˆC and ˆB = ˆD

RTP: ABCD is a parallelogram, i.e. AB || DC and AD || BC

Proof: Let Â = ˆC = x and ˆD = ˆB = y

then Â + ˆB + ˆC + ˆD = 360º . . . sum of the øs of a quadrilateral

â 2x + 2y = 360º

÷ 2) â x + y = 180º

i.e. Â + ˆD = 180º and Â + ˆB = 180º

â AB || DC and AD || BC . . . co-interior øs are supplementary

â ABCD is a parallelogram . . . both pairs of opposite sides ||

DON'T EVER MEMORISE THEOREM PROOFS!

Develop the proofs/logic for yourself before checking against the methods shown below.

Note: We used the result in theorem 2 in the proof of theorem 3 –

but, we could've started from the beginning,

i.e. from the definition of a parallelogram.

We would just have needed to prove an extra pair of Δs

congruent (as in theorem 2).

Theorems: Definition � Property

Converse theorems: Property � Definition

Make sense of

THE LOGIC!

RTP: Required to prove

We could, of course, also have proved the first theorem this way!

It doesn’t matter which

diagonal you draw

1

A B

C D

4

>

>

23

A B

C D

C D

4

1

2

3

>

>

A B

A

>

> B

D

x

y

C

x

y


� Theorem 5: If a QUADRILATERAL has 2 pairs of opposite

sides equal, then the quadrilateral is a ||m Given: Quadrilateral ABCD with AB = CD

and AD = BC RTP: ABCD is a parallelogram,


Construction: Draw diagonal AC . . . it doesn’t matter which diag. you draw

Proof: In Δs ACD and CAB

1) AC is common

2) AD = BC . . . given

3) CD = AB . . . given

â ΔACD ≡ ΔCAB . . . SSS

â ˆ1 = ˆ2 and ˆ3 = ˆ4

â AB || DC and AD || BC . . . alternate øs are equal


� Theorem 6: If a QUADRILATERAL has 1 pair of opposite

sides equal and ||, then the quadrilateral is a ||m Given: Quadrilateral ABCD with AB = and || DC

RTP: ABCD is a parallelogram,


Construction: Draw diagonal AC . . .

Proof: In Δs ABC and CDA

1) AB = DC . . . given

2) ˆ1 = ˆ2 . . . alternate øs; AB || DC

3) AC is common

â ΔABC ≡ ΔCDA . . . SøS

â ˆ3 = ˆ4

â AD || BC . . . alternate øs equal But AB || DC . . . given


� Theorem 7: If a QUADRILATERAL has diagonals which

bisect one another, then the quadrilaterals

is a ||m. Given: Quadrilateral ABCD with diagonals AC and BD intersecting at O and

AO = OC and BO = OD.

RTP: ABCD is a parallelogram,


Proof: In Δs AOB and COD

1) AO = OC . . . given

2) ˆ ˆ1 = 2 . . . vert opp øs

3) BO = O . . . given

â ΔAOB ≡ ΔCOD . . . SøS

â ˆˆBAD = OCD

â AB || DC . . . alternate øs equal

Similarly, by proving ΔAOD ≡ ΔCOB it can be shown that AD || BC

â ABCD is a parallelogram . . . 2 pairs of opp. sides are ||

In our sums, we may use ALL properties and theorem statements . . .

To prove that a quadrilateral is a ||m we may choose one of 5 ways:

1) Prove both pairs of opposite sides || (the definition).

2) Prove both pairs of opposite sides = (a property).

3) Prove 1 pair of opposite sides = and || (a property).

4) Prove both pairs of opposite angles = (a property). . . . THE ANGLES

5) Prove that the diagonals bisect one another (a property). . . . THE DIAGONALS

Using diagonals . . .

To prove a parallelogram is a rectangle: prove that the diagonals are equal.

To prove a parallelogram is a rhombus: prove that the diagonals intersect

at right angles, or

prove that the diagonals bisect

the angles of the rhombus.

. . . THE SIDES

A B

C D

4

1

2

3

It doesn’t matter which

diagonal you draw

A B

C D

3

1

2

4

>

>

A B

C D

2

3

4

1

In Geometry, we never have

to repeat a 'logic sequence'

(as would’ve been required

here) –

we just say: Similarly, . . . !


� Area of Quadrilaterals and Triangles

A SUMMARY OF FORMULAE FOR

AREAS OF QUADRILATERALS

So far, we have established:

� The area of a trapezium = 12

(a + b) .h

� The area of a kite = 12

the product of the diagonals

But also, remember:

� The area of a square = s2,

where s = the length of a side of a square

� The area of a rectangle = ℓ % b

where ℓ = the length & b = the breadth

� The area of a parallelogram = base % height

� The area of a rhombus = . . . . . . . (1) . . . . . . . . . . . . .

or = . . . . . . . (2) . . . . . . . . . . . . .

Answers

(1) base % height (because a rhombus is a parallelogram), or

(2) 12

product of the diagonals (because a rhombus is a kite)

Why is the area of a parallelogram = base % height?

Compare ||m ABCD and rectangle EBCF

||m ABCD = ABCF + Δ2

& rectangle EBCF = ABCF + Δ1

But, we know that the opposite sides of rectangles and

parallelograms are equal â Δ1 Δ2 . . . SS90º or RHS

â Δ1 = Δ2 in area

â Area of ||m ABCD = Area of rectangle EBCF

= BC .FC . . .

= base % height (of the ||m)

ℓ % b

b

ℓ

s

a >

> b

h

NB:

Study THE SUMMARY OF QUADRILATERALS for

'pathways of definitions, areas, properties', etc.

KNOW THIS WELL ! (page 9)

See the explanation of

this formula below: . . .

A

B C

D E F

Δ1 Δ2


SOME IMPORTANT FURTHER FACTS ON AREAS

OF ΔS & QUADRILATERALS

1. Δs on the same base & between the same || lines

are equal in area

Area of Δ = 1

2base (BC) % height (h)

â ΔPBC = ΔABC = ΔQBC in area

2. Parallelograms on the same base and between

the same || lines are equal in area

Area of ||m = base (DC) % height (h)

â ||m ABCD = ||m PQCD in area

3. The median of a Δ bisects the area of the Δ

Area of ΔABD = 1

2BD.h &

Area of ADC = 1

2DC.h

But BD = DC . . .

â Area of ABD = Area of ADC [1

2Area of ABC!]

4. The diagonal of a parallelogram (or rhombus or

rectangle or square!) bisects the area

ΔABC ≡ ΔCDA . . . (SøS)

â ΔABC = ΔCDA [= 1

2 ||m ABCD] in area

5. If two triangles lie between

the same parallel lines:

THE RATIO OF THEIR AREAS

= THE RATIO OF THEIR BASES

area of ABD

area of DAC

Δ

Δ =

1

2

1

2. . h

. 4 . h

x

x

= 1

4 Also,

area of ABD

area of

Δ

ΔABC =

1

2

1

2. . h

. . h

x

x5

= 1

5

base of ABD

= !base of DAC

⎛ ⎞Δ⎜ ⎟

Δ⎝ ⎠ â Area of ΔABD =

1

5area of ΔABC

â Area of ΔABD = 1

4Area of ΔDAC

The height of these Δs & ||ms is the distance between the parallel lines . . .

A

B C D 4x x

h

x x

A

B C D

h

h P

A Q

B

C

h P

A

Q

B

C D

median AD bisects

the base

A B

C D


Gr 10: THE MIDPOINT THEOREM

FACT 1

The line segment through the midpoint of one side of a triangle, parallel to a second

side, bisects the third side.

FACT 2

The line segment joining the midpoints of two sides of a triangle is parallel to the

third side and equal to half of the third side.

PROOFS

1.

2.

Given:

P & Q midpoints of AB & AC Result:

PQ || BC & PQ =1

2BC

P Q

C B

A

Given:

P midpoint AB & PQ || BC Result:

Q midpoint AC & PQ =1

2BC

P Q

C B

A

(See Exercise 4 Q3.2 for the proof)

(See Exercise 4 Q3.1 for the proof)

Regard these Facts 1 & 2 as a special case

of the Proportion Theorem in Gr 12 Geometry.

Use the diagrams below to prove facts 1 and 2:

Grade 8 to 10

EUCLIDEAN GEOMETRY

EXERCISES & FULL SOLUTIONS

EXERCISE 1: Lines and Angles

EXERCISE 2: Lines, Angles and Triangles

EXERCISE 3: Quadrilaterals

EXERCISE 4: Midpoint Theorem


EXERCISE 1: Lines and Angles

(Answers on page 22)

1. 140º and 40º are adjacent supplementary angles:

The angle supplementary to x is :

2. In this figure : x + 90º + y = 180º . . . øs on a straight line

â x + y = and so they are called angles. 3. Is ADB a straight line? Give a reason for your answer.

4.

5. x + y = 180º . . . øs on a

straight line

and z + y = 180º . . . øs on a

straight line

â

6. Given: reflex ˆ

AOD = 200º (see figure below). Reasons:

Obtuse ˆ

AOD =

ˆ

BOD = ˆ

AOC = 7.

ˆ

1 and ˆ2 are ˆ

1 and ˆ2 are ˆ

1 and ˆ2 are ø

s (NAME) øs (NAME) ø

s (NAME)

RELATIONSHIP: RELATIONSHIP: RELATIONSHIP:

8. ˆ

1 and are corresponding øs and they are

ˆ

1 and are alternate øs and they are

ˆ

1 and are co-interior øs and they are

9. NB: It is ONLY BECAUSE THE LINES ARE , that

the corresponding and alternate øs ARE EQUAL and the

co-interior øs are SUPPLEMENTARY !!!

40º 140º

y x

(2.1) (2.2)

145º 45º

A D B

C

(3.1) (3.2) x90º - x

A D B

CE

100º 40º

60º ?

(4.1) x ?

(4.2)

The sum of the

adjacent angles

about a point

is 360º.

When two lines intersect,

the vertically opposite øs

are equal. Why?

(5)

O A

C

B

D

200º (6.1)

(6.3) (6.4)

(6.5) (6.6)

1

2

1

2

(7.1)

1

2

(7.2) (7.3)

(7.4) (7.5) (7.6)

1

2

3

4

5

6

(8.1) (8.2)

(8.3) (8.4)

(8.5) (8.6)

(9)

(1) ? x

(6.2)

x

zy


EXERCISE 2:

Lines, Angles and Triangles


1. In the sketch,

AB is a straight line.

If x - y = 10º, find the

value of x and y.

2.1 Find the size of angles a to g (in that order), giving reasons.

2.2 Calculate x and give reasons.

2.2.1 2.2.2 2.3 If PQ = SR and

øPQR = øSRQ,

prove that PR = SQ.

(Hint : first prove that ∆PQR ≡ ∆SRQ) 2.4 State whether PQ || RS,

giving reasons.

3. Determine the values of x and y in the following diagrams. Give reasons for your answers.

3.1 3.2

4. Prove that �P = ˆT

by first proving that

the 2 triangles

are congruent.

5. State whether the following pairs of triangles are

congruent or similar, giving reasons for your choice. 5.1 5.2

5.3

6.1 Find the value of x, by forming an equation first

and then solving for x.

Show all your working and give reasons. 6.1.1 6.1.2 6.2 In the diagram,

AB || ED and BE = EC.

Also, ˆABE = 27º

and ˆBAC = 53º.

6.2.1 Write down the sizes of ˆBED and ˆCED ,

giving reasons. 6.2.2 Hence, or otherwise, calculate the

value of x, showing all working and

giving reasons. 7. Calculate the area

of the kite alongside.

x + yy x

g

b c d

a

ef

35º 60º

3x 4x

5x

P R

T W

Q S

76º

VU 104º

B

A C

56º

37º

18

30

X

Z Y 30

18

87º

N

69

L

M 75

51H

2317

25

E

F

D

CB A

3x x

66º

A

DC

B 3x - 10º

x + 30º

F A C

E

G B D

3xx

y 60º

2x

120º

110º

x

P S

RQ

P

T

Q R

Sx

x

A

B D C

53º

27º

x

E

4 cm

12 cm

13 cm

A

C

B D E

K

T R

P

Q A C

E

F

B D

y 100º

60º

x

120º


8. Calculate the value

of x giving

reasons.

9. If AC || DB, prove with

geometric reasons that

x = y + z.

10. In ΔABC, BC is produced to D and CE || BA.

Prove that:

10.1 ˆˆ ˆ

1A + B + C = 180º

10.2 ˆ ˆ ÂCD = A + B

11. Prove that ˆDOC = 45º.

12. If AB || CD,

ˆBOE = 140º and

ÂOC = 35º. Determine,

stating reasons, the values of

x and y.

13. In the accompanying

figure BG || CF.

Complete the

following statements

(giving reasons):

13.1 ˆ ˆ ˆ ˆ1 + 2 + 3 + 4 = . . . . (degrees)

13.2 �ˆ1 + 11 = . . . . (degrees)

13.3 ˆ ˆˆ6 + 7 + 2 = . . . . (degrees)

13.4 ˆ ˆ9 + 6 = . . . (the no. of one angle)

13.5 ˆ7 = . . . . (list only one angle)

13.6 If � �12 = 10 , then GD = . . . .

14.1 Express ˆCEB , ÂBE and

ÂEB in terms of x. 14.2 What is the relationship

between CE and AD? Explain.

14.3 If CE bisects ÂCD

what is the relationship between BE and CD? Explain.

15.1 Make a neat copy of this sketch and fill in all the

other angles in terms of x. Reasons are not required.

15.2 Complete the following statement:

∆ABE ||| ∆ . . . ||| ∆ . . . 15.3 If BC = 18 cm and BE = 12 cm,

calculate the length of 15.3.1 AE 15.3.2 AB correct to one decimal. 15.4 Hence calculate the area of rectangle ABCD.

16. In each of the following, state whether the given triangles are congruent or not , and in each case give a reason for your answer. Do not prove the triangles congruent, but name each congruent pair correctly.

16.1 16.2 16.3 16.4 16.5

17. Given ΔPQR and ΔABC.

17.1 Show that ΔPQR is NOT similar to ΔABC.

Clearly show all relevant calculations

and reasons. 17.2 Prove that ΔPQR is not right-angled.

A

2x + 36º

B C D

3x - 50º

x +10º

A

C

y D

B

z

x

A B C D E

5

F G

H

10

12

6789

2

31

4

11

A E D

CB

x

SP

Q R

A B

C

D E

B is the centre of the circle

B

A

C

D

B

A

D C

A

9 6

3 B C

P

R

6 Q

812

EA

B C D

y

y

x

x

A

B C D

E

231

B

A D C

O y y

xx

B E

D C

A

x

A B

C D

E

x

35º 21

3140º

1

y

O




1. Which of the following quadrilaterals is definitely a

parallelogram? (Not drawn to scale.) A. B. C. D.

2. If the area of

parallelogram PQRS

is equal to 90 cm2

then PT is equal to

. . . cm.

3. In the given diagram,

PQRS is a parallelogram.

SP = PT and

UR = RQ.

Prove that TRUP is a parallelogram.

4. In the diagram,

ABCD is a

parallelogram.

AB is produced

to E and AD to F. Prove that ECF

is a straight line.

5. ABCD is a rhombus. By setting up an equation

and showing all steps and

reasons in the process,

find the value of y.

6. ABCD is a parallelogram. E is a point on AD such that AE = AB and EC = CD.

ˆBEC = 90º.

Calculate the size of ÊBC .

7. In each of the following cases, calculate the value

of x giving reasons. 7.1 WXYZ is a

parallelogram. 7.2 ABCD is a square.

ˆBFD = 125º.

8. Prove that ABCD is a parallelogram. 9. In the accompanying

figure ABCD is a

parallelogram.

The diagonal is

produced to E and

AB = BE and

AD = CE.

If ˆCEB = x, prove, giving reasons, that ˆFDC = 3x.

10. LM = 5 cm,

LP = 6 cm and

PS = 1 cm.

What type of quadrilateral is LMPN? Give reasons.

Find NM and MS.

11. ABCD is a parallelogram, and

AE and BF bisect Â and ˆB

respectively.

P is the point of intersection

of AE and BF. 11.1 Find the magnitude

of ÂPB . 11.2 Show that BC = CF. 11.3 Prove that DF = EC.

12. ABCD is a parallelogram.

AP bisects ˆDAB and

BP bisects ÂBC . Prove that AB = 2BC

13.1 Prove the theorem which says that if both pairs of

opposite angles of a quadrilateral are equal,

then it is a parallelogram.

13.2 PQRS is a quadrilateral

with PS = PR = QR and

PQ || SR.

Prove that it is a parallelogram. 13.3 Write down 2 facts about a rhombus which are

not generally true of any parallelogram.

14. PQRS is a parallelogram.

Prove that ABCD is a rectangle.

P S

Q T R

15 cm

10 cm

P Q

S T R

U

A B

CD4y - 18º

y

A

125º

x

B

CD

F

E

A B

CD

E x

F

A B

CD E F

x yx

P

y

B C

A DE

W

3x - 30º

x + 10º

X

Y Z

A B

CD P

1 2

2 1 3

1

2

P

S

C

Q

R

D

A

B

m m x

x

n n

y y

A D F

C

E

B

P Q

S R

1 2

L

NP

M

S

R

A D

B F C

E

65º

65º

115º


P

Q

R

S

W

T V



1.1 Complete (giving missing words only):

The line segment joining the midpoints of two sides

of a triangle is . . . . . . to the third side and

equal to . . . . . . .

1.2 ΔABC has medians

BN and CM drawn,

cutting each other

at O.

P and Q are the

midpoints of BO

and CO

respectively. Prove that MNQP is a parallelogram.

2. In the diagram below QS || TV; PQ || ST;

QT = TR = 9 cm and PS = 15 cm.

2.1 Prove VR = 71

2cm.

2.2 Calculate PQ if PQ =16

5VR and hence

prove that ˆPQR = 90º. 2.3 Write down the length of ST.

3.1 With reference to the diagram, prove the theorem which states that PQ is parallel to DC and half its length.

3.2 The diagram shows a

triangle PQR with M the

midpoint of PQ.

MNT is drawn parallel

to QR so that RT || QP. Use this diagram to

prove the theorem

which states that

MN bisects PR.

4. Triangle DEF has G

the midpoint of DE,

H the midpoint of DF.

GH is joined.

HJ is parallel to DE.

Prove:

4.1 GHJE is a parallelogram.

4.2 ˆGDH = ˆJHF

4.3 JF = GH

5. M, N and T are the

midpoints of AB,

BC and AC in ∆ABC.

ˆA = 60º and ˆB = 80º. Calculate the angles

of ∆MNT.

6. ABCD is a square.

The diagonals AC and

BD intersect at O.

M is the midpoint of BO

and AM = ME. 6.1 Prove that MD || EC. 6.2 Prove that DOEC is

a parallelogram. 6.3 Determine, with reasons,

the size of :

6.3.1 ˆ

1D

6.3.2 ˆECD

7. With reference to the diagram,

prove that 7.1 FE = EC = CA

7.2 ΔBCA ≡ ΔAED

8. In ΔPQR, PT = TQ, while PV = VR = RS.

Show that WR =1

4QR.

9. ABCD is any quadrilateral.

E is a point BC. P, Q, R and S are the midpoints

of AB, AE, DE and DC.

Prove that :

9.1 PQ || RS

9.2 PQ + QR + RS =1

2(AD + BC)

P

S

Q T R

15

9 9

V

A

B CN

T

60º

M

80º

O

M

1

A D

E

CB A

D

P Q

C

D

G H

FJ E

A

B

P Q

E C

D

SR

A

B C

MO

N

P Q

P

MT

R

N

Q

O

F

x

A

B

D

E

y

y x

C


ANSWERS TO EXERCISES

EXERCISE 1: Lines & Angles

1. 180º - x 2.1 90º 2.2 complementary 3.1 No . . . 45º + 145º ≠ 180º 3.2 Yes . . . 90º – x + 90º + x = 180º 4.1 160º 4.2 360º - x 5. â x = z 6.1 160º 6.2 revolution / øs about a point

6.3 20º 6.4 øs on a straight line

6.5 20º

6.6 øs on a straight line or vertically opposite to ˆBOD

7.1 corresponding 7.2 co-interior 7.3 alternate 7.4 equal 7.5 supplementary 7.6 equal 8.1 ˆ5 8.2 equal

8.3 ˆ4 8.4 equal

8.5 ˆ3 8.6 supplementary 9. parallel

EXERCISE 2:

Lines, Angles and Triangles

1. 2x + 2y = 180º . . . øs on a straight line

â x + y = 90 But x - y = 10º â x = 50º and y = 40º . . . by inspection 2.1 a = 60º . . . vertically opposite angles

b = 35º . . . alternate øs ; || lines

c = 35º . . . øs opp = sides

d = 85º . . . sum of øs in ∆

e = 60º - 35º = 25º . . . ext. ø of ∆

f = 35º + 35º = 70º . . . either ext. ø of ∆ or

corresp. øs ; || lines

g = 60º – 35º = 25º . . . ext. ø of ∆ 2.2.1 12x = 360º . . . revolution â x = 30º 2.2.2 120º + 110º + x = 2(180º) â 230º + x = 360º â x = 130º

2.2.3 In ∆s PQR & SRQ [Note: order of letters!]

(1) PQ = SR . . . (given)

(2) ˆPQR = ˆSRQ . . . (given)

(3) QR is common â ∆PQR ≡ ∆SRQ . . . SøS

â PR = SQ

2.4 ˆPUV = 180º - 76º = 104º . . . øs on str line

â ˆPUV = ˆRVW

â Yes, PQ || RS . . . because corresponding

øs are equal

3.1 x = 100º - 60º = 40º . . . ext. ø of EFC

y = 120º - 40º = 80º . . . ext. ø of ABC 3.2 4x = 2x + 60º . . . corr. øs ; BA || DC

â 2x = 60º â x = 30º

y = 180º - 4x = 60º . . . øs on a straight line

4. In ∆s QRP and SRT

(1) ˆQ = ˆS (= x)

(2) ˆQRP = ˆSRT . . . vertically opposite øs

(3) QR = SR . . . given â ∆QRP ≡ ∆SRT . . . øøS

â ˆP = ˆT

5.1 Congruent ; SøS . . . [ ˆC = 87º ø sum of ∆] 5.2 Similar ; sides in proportion

. . . [17:23:25 = 51:69:75] 5.3 Similar ; equiangular (alt. øs & vert. opp. øs )

. . . [no sides]

LOOKS MAY NOT DECEIVE!

6.1.1 3x = 66º + x . . . ext ø of ∆

â 2x = 66º â x = 33º 6.1.2 (3x - 10º) + (x + 30º) = 180º . . . co-int. øs; AB || CD

â 4x + 20º = 180º â 4x = 160º â x = 40º

6.2.1 ˆBED = 27º . . . alt. øs ; AB || ED

ˆCED = 53º . . . corresp. øs ; AB || ED 6.2.2 ÂBD = x . . . corresp. øs ; AB || ED

â ÊBC = x - 27º

â ˆC = x - 27º . . . øs opp = sides â In ∆ABC: 53º + x + x - 27º = 180º . . . sum of øs in ∆ â 2x = 154º â x = 77º 7. BE ⊥ AC . . . diagonals of a kite BE = 5 cm . . . 5:12:13 ∆ ; Pythag.

Area of kite = 2. 1

2(12 + 4).5 . . . 2 % ∆

= 80 cm2

[OR: 12

product of diagonals . . . why?]

8. 2x + 36º = 3x - 50º + x + 10º . . . ext. ø of ∆

â -2x = -76º â x = 38º A VERY IMPORTANT THEOREM!

9. ˆC = z . . . alt. øs ; AC || DB

â x = y + z . . . ext. ø of ∆ 10.1 ˆ

3C + ˆ

2C + ˆ

1C = 180º . . . øs on a straight line But ˆ

2C = Â . . . alternate øs ; CE || BA

& ˆ

3C = ˆB . . . corresp. øs ; CE || BA â ˆB + Â + ˆ

1C = 180º

10.2 ÂCD = 180º - ˆ

1C . . . str. line

& Â + ˆB = 180º - ˆ

1C . . . ø sum of ∆

â ÂCD = Â + ˆB

11. ˆDOC = x + y . . . ext. ø of ∆ But 2x + 2y = 90º . . . ø sum of rt. ød ∆ABC

â x + y = 45º â ˆDOC = 45º

12. ˆ

1O = 40º & ˆ

3O = 40º . . . øs on a straight line

â x = 40º . . . alt. or corresp. øs ; AB || CD

ˆ

1C = 35º . . . alt. øs ; AB || CD

â y = 145º . . . øs on a straight line

. . . 2 prs. co-int. øs;

|| lines

Copyright © The Answer 23

13.1 360º . . . revolution

13.2 180º . . . co-interior øs ; BG || CF

13.3 180º . . . vert. opp. øs ; ø sum of ∆

13.4 �12 . . . ext. ø of ∆

13.5 ˆ9 . . . corresp. øs ; BG || CF

13.6 BD . . . �

12 = �

10 � �11 = �9 ; sides opp = øs 14.1 ˆCEB = x . . . øs opp = sides

ÂBE = 2x . . . ext. ø of ∆

ˆÂ + AEB = (180º - 2x) . . . sum of øs in ∆

â ÂEB = 90º - x . . . øs opp = sides

14.2 They are perpendicular to each other; i.e. CE ⊥ AD;

Explanation: ˆCEB and ÂEB are complementary

. . . x + (90º - x) = ? 14.3 They are parallel; i.e. BE || CD ;

Explanation: ÊCD = x . . . CE bisects ÂCD

â ÊCD = alternate angle, ˆCEB

15.1 15.2 ∆ABE ||| ∆ECB ||| ∆DEC 15.3.1 ∆ABE ||| ∆ECB

� AEBE

= BE

BC . . . sides in proportion

%BE) â AE = 2 2

BE 12 = = 8 cm

BC 18

15.3.2 AB2 = 122 - 82 = 80 . . . Theorem of Pythagoras

â AB = 80

j 8,9 cm 15.4 Area of rect. ABCD = 8,9 % 18 j 161 cm2

16.1 Yes, ∆ABC ≡ ∆EDC . . . øøS 16.2 No . . . 2 sides and an angle, but the angle isn’t included 16.3 Yes, ∆ABD ≡ ∆ACD . . . RHS

16.4 Yes, ∆ADB ≡ ∆CDB . . . SSS [Note: equal radii] 16.5 No . . . 2 angles and a side, but they don’t correspond

17.1 12 4 8 4 6= ; = but = 2

9 3 6 3 3

â The sides are not in proportion

â ∆PQR ||| ∆ABC

17.2 122 = 144 and 62 + 82 = 36 + 64 = 100

â 122 g 62 + 82

â ˆQ g 90º, i.e. ∆PQR is not rt. ød. . .


1. D: diagonals bisect one another

2. Area of ||m PQRS = 15 % PT = 90 . . . base % height

â PT = 6 cm 3.

Let ˆS = x

Then ˆPTS = x . . . øs opp = sides

â ˆTPQ = x . . . alt. øs ; PQ || SR in ||m PQRS

But ˆQ = x . . . opp. øs of ||m PQRS

â ˆRUQ = x . . . øs opp = sides

â ˆTPQ = ˆRUQ

â PT || UR . . . corresponding øs equal But PU || TR . . . opposite sides of ||m PQRS

â TRUP is a parallelogram . . . both pairs opp. sides ||

4.

Let ˆF = x

then ˆBCE = x . . . corr. øs ; BC || AD(F) in ||m Similarly, let Ê = y

then ˆDCF = y In ΔAEF: Â + x + y = 180º . . . ø sum of

But Â = ˆBCD . . . opposite øs of ||m

â ˆBCD + x + y = 180º

â ECF is a straight line . . . conv. of 'øs on a str. line'

5. ˆDAC = y . . . diagonals of a

rhombus bisect the

øs of the rhombus

â ˆDCA = y . . . øs opp = sides,

sides of a

rhombus (or alternate angles) â 4y - 18º + 2y = 180º . . . ø sum of

â 6y = 198º â y = 33º 6. x1 = x2 . . . øs opp = sides

= x3 . . . alt. øs ; AD || BC in ||m

(= x, say)

Similarly, y1 = y2 = y3 (= y, say) x + y = 90º . . . str. ÂED or øs of BEC

and y = 2x . . . opp. øs of ||m

â 3x = 90º

â x = 30º, i.e. ÊBC = 30º

7.1 3x - 30º = x + 10º . . . opp. øs of a ||m

â 2x = 40º â x = 20º

B E

D C

A

x

2x x

P Q

S T R

U x

x

x

x

x

A B

CD 4y - 18º

y y

y

W

3x - 30º

x + 10º

X

Y Z

A D

x1

x3

x2 y2 y1

y3

B C

E

The angle is not

included, but

it is a right angle

CONVERSE

of Theorem

of Pythag.

A E D

C B

x

90º- x

12

x

90º-x

90º- x 18

x

NB: The order of

the letters!

A D F

C

E

B

y

y

x

x


7.2 ˆBAE = 45º . . . diagonals of a square bisect

(right) angles of a square

â ÂBE = x - 45º . . . ext. ø of Δ ˆFAB = 90º . . . ø of square ˆBFD = ÂBE + ˆFAB . . . ext. ø of Δ

â 125º = x - 45º + 90º â x = 80º

8. ÂEC + ˆFCE = 115º + 65º = 180º

â AE(D) || (B)FC . . . co-interior øs supplementary ˆB = 65º . . . øs opp = sides & ˆCED = 65º . . . str. line AED

â ˆD = 65º . . . øs opp = sides

â ÊCD = 50º . . . ø sum of Δ â ˆB + ˆDCB = 65º + 50º + 65º = 180º

â AB || DC . . . co-interior øs supplementary

â ABCD is a ||m . . . 2 pairs opp. sides || 9.

ÊAB = x . . . øs opp = sides

â ˆDCA = x . . . alt. øs ; DC || AB in ||m ˆCBE = x . . . øs opp = sides

â ÂCB = 2x . . . ext. ø of Δ

â

ˆDAC = 2x . . . alt. øs ; AD || BC in ||m ˆFDC = ˆDAC + ˆDCA . . . ext. ø of Δ

= 2x + x

= 3x

10. A rhombus ; 2 prs. opp. sides || and diagonals intersect

at right øs

LR = RP = 3 cm . . . diagonals bisect

â RM = 4 cm . . . ˆLRM = 90º ; 3:4:5∆ ; Pyth.

â NM = 8 cm . . . diagonals bisect

MP = 5 cm . . . sides of rhombus

â MS2 = 52 - 12 = 24 . . . Pythagoras

â MS = 24 j 4,9 cm

11.1

2x + 2y = 180º . . . co-int. øs ; AD || BC in ||m

â x + y = 90º

â ÂPB = 90º . . . ø sum of

11.2 ˆBFC = ÂBF = y . . . alt. øs ; AB || DC

âˆCBF = ˆCFB

â BC = CF . . . sides opp = øs

11.3 DF = DE - FE & EC = CF - FE ÂED = ˆBAE (= x) . . . alt. øs ; AB || DC in ||m

â DE = AD . . . øs opp = sides

= BC . . . opp. sides of ||m = CF . . . proved in 11.2 â DF = EC

12.

ˆ

1P = ˆ

2A . . . alt. øs ; AB || DC in ||m

= ˆ

1A (= x) . . . given

â DP = DA . . . øs opp = sides Similarly, ˆ

3P = ˆ

1B = ˆ

2B = y

â CP = BC But BC = DA . . . opp. sides of ||m

& AB = DC . . . opp. sides of ||m = DP + CP = DA + BC = 2BC 13.1 In the notes (Theorem 4)

13.2 PQ || SR Let ˆS = x; then

ˆPRS = x . . . øs opp = sides

â ˆ

2P = x . . . alt. øs ;

PQ || SR

â ˆQ = x . . . øs opp = sides

â In ∆s PSR and PRQ

ˆ

1P and ˆPRQ = 180º - 2x . . . ø sum of s

â PS || QR . . . alt. øs equal

â PQRS is a ||m . . . 2 prs. opp. sides || OR: Could've proved 2 prs. opp. øs equal

OR: Could've proved 2 prs. opp. sides equal

13.3 The diagonals intersect at right angles. The diagonals bisect the angles of the rhombus. (& 2 adjacent sides are equal)

L

N P

M

S

R

5

3

3

1

A B

C D EF

x y x

P

y

y x

A B

C D

EF

2x

2x x

x

x

x

x

x

y

y

y

A B

C D P

1 2

2 1 3

1

2

P Q

S R

1 2

x x

x x

A

125º

x

B

CD

F

E

65º

A D

B F C

E

65º

65º

115º

65º

65º

50º

Copyright © The Answer 25

14.

2m + 2n = 180º . . . co-int. øs ; PQ || SR in ||m

â m + n = 90º

â ˆPAS = 90º . . . ø sum of

â ˆBAD = 90º . . . vertically opp øs Similarly, x + y = 90º

â ˆQCR = 90º

â ˆBCD = 90º So, too, n + y = 90º

â ˆB = 90º . . . ø sum of â The 4th angle, ˆD = 90º . . . ø sum of quad. â ABCD is a rectangle . . . all angles = 90º


1.1 . . . parallel . . . half of the third side. 1.2

Join MN and PQ

In ∆ABC: M & N are midpoints of AB & AC

â MN || BC and MN = 12

BC . . . midpoint thm

& In ∆OBC: P & Q are midpoints of OB & OC

â PQ || BC and PQ = 12

BC . . . midpoint thm

â MN || PQ and MN = PQ . . . both || and =1

2BC

â MNQP is a ||m . . . 1 pair of opp. sides = and ||

2.1 In ∆RQP: T midpoint QR & TS || QP â S midpoint PR . . . â SR = 15 cm Similarly : In ∆RQS: V midpoint SR

â VR = 12

(15) = 17 cm

2

2.2 PQ = 16 15= 24 cm

5 2%

â sides of ∆PQR:

18:24:30 = 3 :4:5

â ˆPQR = 90º . . . ratio of sides = Pythag. 'triple'

i.e. converse Pythag.

2.3 ST = 12

(24 cm) = 12 cm . . .

3.1 Construction:

Extend PQ to R

such that PQ = QR

Join AR, PC and CR

Proof: APCR is a ||m . . . diagonals bisect one another

â CR = and || AP

â CR = and || PD

â PDCR is a ||m . . . 1 pr. opp. sides = and ||

â PR || and = DC

â PQ || and = 12

DC . . . PQ = 1

2PR

3.2. MQRT is a ||m . . .

â TR = and || MQ

â TR = and || PM

â In ∆s PNM & RNT (1) PM = RT . . . proved

(2) ˆPNM = ˆRNT . . . vertically opp øs

(3) ˆPMN = ˆNTR . . . alternate øs ; PM || TR

â ∆PNM ≡ ∆RNT . . . øøS

â PN = NR

4.1

HJ || GE . . . given & In ∆DEF: G & H are midpoints of DE & DF

â GH || EJ(F) [and GH = 12

EF]

â GHJE is a ||m . . . 2 pairs opp. sides ||

4.2 ˆGDH = ˆJHF . . . corresp. øs ; HJ || DE

4.3 EJ = GH . . . opp. sides of ||m

= 1

2EF . . . see 4.1

(â JF = 1

2EF too)

â JF = GH 5.

ˆC = 180º - (80º + 60º) . . . sum of øs in

= 40º In ΔABC: M & T are midpoints of AB and AC

â MT || BC . . . midpoint thm.

â ˆAMT = 80º . . . corresp. øs ; || lines Similarly, M & N are midpoints of AB & BC

â ˆBMN = 60º . . . corresp. øs ; MN || AC

â ˆTMN = 180º - (80º + 60º) . . . øs on a str. line

= 40º The same method can be followed to determine the

other two angles of ΔMNT. Answer: 40º ; 80º ; 60º

OR: Pythag

9 :12 : 15

= 3 :4 :5 !

midpoint

theorem

converse of

midpoint theorem

A

B C

M O

N

P Q

A

D

P Q

C

R A

B C N

T

60º

M

80º

P

S

C

Q

R

D

A

B

m

m x

x

n

n y

y

P

S

Q T R

15

9 9

V

2 pairs

opp. sides ||

D

G H

F J E

P

M T

R

N

Q


P

Q R

S

W

T V

6.1 In ∆AEC: O midpoint AC . . . diagonals bisect

& M midpt. AE . . . given â MO(D) || EC . . . midpoint theorem

6.2 We have DO || CE . . . in 6.1

& DO = OB . . .

= 2OM . . . M midpt. BO

= EC . . . midpoint. theorem â DOEC is a ||m . . . 1 pair of opp. sides = and ||

6.3.1 ˆ

1D = 45º . . . right øs of sq. bisected by diagonal

6.3.2 ÊCA = ˆBOA . . . corresp. øs ; MO || EC in 6.1

= 90º . . . diagonals of sq. int. at rt. øs

& ÔCD = 45º . . . just as ˆ

1D above

â ÊCD = 90º + 45º = 135º

7.1 In ∆FBC: D midpoint FB & DE || BC

â FE = EC . . . converse of midpoint thm & In ∆EDA: O midpoint AD and (B)OC || DE

â EC = CA . . . converse of midpoint thm â FE = EC = CA

7.2 In ∆s BCA & AED (1) ˆCBA = ÊAD . . . both = x

(2) ˆBCA = ÂED . . . corresp. øs ; DE || BC

(3) CA = FE . . . in 7.1

= ED . . . øs opp = sides

â ∆BCA ≡ AED . . . øøS

8. In ∆PQR: T & V are midpoints of PQ & PR

â TV || QR and = 12

QR . . .

â In ∆STV: R midpoint VS and WR || TV

â WR = 12

TV

= 12

( 1

2QR)

= 14

QR

9.

9.1 In ΔABE:

P & Q are midpoints of AB & AE

â PQ || BE(C) Similarly, in ΔDEC:

RS || (B)EC

â PQ || RS . . . both are

parallel

to BEC

9.2 In Δs ABE, AED and DEC:

PQ + QR + RS = 12

BE + 12

AD + 12

EC

= 12

(AD + BE + EC)

= 12

(AD + BC)

midpoint

theorem

These Geometry materials (Booklets 1 to 4)

were created and produced by

The Answer Series Educational Publishers (Pty) (Ltd)

to support the teaching and learning of

Geometry in high schools in South Africa.

They are freely available to anyone

who wishes to use them.

This material may not be sold (via any channel)

or used for profit-making of any kind.

This drawing looks confusing at first. But, look at each

triangle separately – the 'middle' one is just upside down!

– and apply the facts to each, one at a time.

A

B

P Q

E C

D

SR

diagonals of

sq. bisect

O

M

1

A D

E

CB

O

F

x

A

B

D

E

y

y xC


Circle Geometryby The TAS Maths Team

Circle Geometry



WWW.THEANSWER.CO.ZA


CIRCLE GEOMETRY

The Language (Vocabulary)

GROUP AND

� Centre

� Diameter

� Radius � Circumference

� Chords

� Arcs (major & minor)

� Segments (major & minor)

� Sectors

� "SUBTEND" . . . Understand the word!

� Central and Inscribed angles

In all the figures, arc AB �(AB), or chord AB, subtends:

� a central ˆAOB at the centre of the circle, and

� an inscribed ˆAPB at the circumference of the circle.

A

B O

C

diameter

radius

centre

B

A

chord

B A

minor segment

major segment

sector B

A

O

Consider that

subtend

means support.

1 2 Figure 2 Figure 1

P

A B

O

Figure 3

P

A B O

180º

P

A B

O

Figure 4

P

A

B

O

B

A

major arc AB

chord AB

minor arc AB

To ensure that you grasp the meaning of the word 'subtend' :

� Take each of the figures:

� Place your index fingers on A & B;

� move along the radii to meet at O and back; then,

� move to meet at P on the circumference and back.

� Turn your book upside down and sideways.

You need to recognise different views of these situations.

� Take note of whether the angles are acute, obtuse, right, straight or reflex.

� Redraw figures 1 to 4 leaving out the chord AB completely and observe

the arc subtending the central and inscribed angles in each case.

P

O


GROUP

� Cyclic Quadrilaterals

A cyclic quadrilateral is a quadrilateral which has all 4 vertices on the circumference

of a circle.

Note: Quadrilateral AOCB is not a cyclic quadrilateral because point O

is not on the circumference! (A, O, C and B are not concyclic)

� Exterior angles of polygons

The exterior angle of any polygon is an angle which is formed between one side of

the polygon and another side produced. e.g. A triangle e.g. A quadrilateral /

cyclic quadrilateral

ˆACD is an exterior ø of ΔABC. ˆADE is an exterior ø of c.q. ABCD.

[NB: BCD is a straight line! ] [NB: CDE is a straight line! ]

GROUP

� Tangents

Special lines

� A tangent is a line which touches a circle at a point.

� A secant is a line which cuts a circle (in two points).

3 4

NB: It is assumed that the

tangent is perpendicular

to the radius (or diameter)

at the point of contact.

Points A, B, C and D

are concyclic,

i.e. they lie on the same circle. A

C

O

B

D

A

B

D

C

A

C

B D

E

point of contact

tangent

secant

We name quadrilaterals by going around, either way, using

consecutive vertices, i.e. ABCD or ADCB, not ADBC.


2x

x x

x

2x

x + y = 180º

y

x


SUMMARY OF CIRCLE GEOMETRY THEOREMS

TThhee

''CCeennttrree''

ggrroouupp

TThhee

''TTaannggeenntt''

ggrroouupp

TThhee


ggrroouupp

TThhee


ggrroouupp

Equal

radii !


a cyclic quad '.

III

I

IV

II

Equal

chords!

Equal

tangents!


2x

x x

x

2x

x + y = 180º

y

x


GROUPING OF CIRCLE GEOMETRY THEOREMS

The grey arrows indicate how various theorems are used to prove subsequent ones

TThhee

''CCeennttrree''

ggrroouupp

TThhee

''TTaannggeenntt''

ggrroouupp

TThhee


ggrroouupp

TThhee


ggrroouupp

Equal

radii !


a cyclic quad '.

III

I

IV

II

Equal

chords!

Equal

tangents!


and, vice versa, equal angles are subtended

by equal chords.

Copyright © The Answer Series i

CIRCLE GEOMETRY THEOREMS

PAPER 2: GEOMETRY

� Circle Geometry Theorems

Given: ?O with OP ⊥ AB

To prove: AP = PB

Construction: Join OA and OB

Proof: In Δs OPA & OPB

(1) OA = OB . . . radii

(2) ˆ

1P = ˆ

2P (= 90º) . . . given

(3) OP is common

â ΔOAP ≡ ΔOBP . . . RHS

â AP = PB, i.e. OP bisects chord AB � . . .

Given: ?O with AP = PB

To prove: OP ⊥ AB

Construction: Join OA and OB

Proof: In Δs OPA & OPB

(1) OA = OB . . . radii

(2) AP = PB . . . given

(3) OP is common

â ΔOAP ≡ ΔOBP . . . SSS

ˆ

1P = ˆ

2P

But, ˆ

1P + ˆ

2P = 180º . . . øs on a straight line

â

ˆ

1P = ˆ

2P = 90º

i.e. OP AB �

Given: ?O, arc AB subtending ÂOB at the centre and ÂPB at the circumference.

To prove: ÂOB = ˆ2APB

Construction: Join PO and produce it to Q.

Proof: Let ˆ

1P = x

Then Â = x . . . øs opposite equal radii

â ˆ

1O = 2x . . . exterior ø of ΔAOP

Similarly, if ˆ2

P = y, then ˆ

2O = 2y

â ÂOB = 2x + 2y

= 2(x + y)

= 2 ÂPB �

Given: ?O and cyclic quadrilateral ABCD

To prove: Â + ˆC = 180º & ˆB + ˆD = 180º

Construction: Join BO and DO.

Proof: Let Â = x

Then ˆ1

O = 2x . . .

â ˆ

2O = 360º – 2x . . . øs

about point O

â ˆC = 1

2(360º – 2x) = 180º – x . . .

â Â + ˆC = x + 180º – x = 180º

& â ˆB + ˆD = 180º � . . .

1 2

AP

B

O

Q

P

A

B

y

y

x

x

1

1 2

2

O

A

O

B

1

C

D

2

x

ø at centre = 2 %

ø at circumference

ø at centre = 2 %

ø at circumference

sum of the øs of a

quadrilateral = 360º

1 2 A

P B

O

1 The line segment drawn from the centre of a circle,

perpendicular to a chord, bisects the chord.

2 The line drawn from the centre of a circle that

bisects a chord is perpendicular to the chord.

The angle which an arc of a circle subtends at the centre is

double the angle it subtends at any point on the circumference. 3

The opposite angles of a cyclic quadrilateral are supplementary. 4 This proof has

been added in the

2021 Exam Guidelines.

Copyright © The Answer Series ii

Method 1

Given: ?O with tangent at N and

chord NM subtending ˆK

at the circumference.

RTP: ˆMNQ = ˆK

Construction: radii OM and ON

Proof: Let ˆMNQ = x

ÔNQ = 90º . . . radius ⊥ tangent

∴ ÔNM = 90º – x

∴ ÔMN = 90º – x . . . øs opposite equal radii

∴ ˆMON = 2x . . . sum of øs in Δ

∴ ˆK = x . . . ø at centre = 2 % ø at circumference

â ˆMNQ = ˆK �

Method 2

Given: ?O with tangent at N and

chord NM subtending ˆK

at the circumference.

RTP: ˆMNQ = ˆMKN

Construction: diameter NR; join RK

Proof: ˆRNQ = 90º . . . tangent ⊥ diameter

& ˆRKN = 90º . . . ø in semi-?

Then . . .

Let ˆMNQ = x

â ˆRNM = 90º – x

â ˆRKM = 90º – x . . . øs in same segment

â ˆMKN = x

â ˆMNQ = ˆMKN �

The angle between a tangent to a circle and a

chord drawn from the point of contact is equal to the

angle subtended by the chord in the alternate segment. 5

These proofs are

logical & easy to follow.

Draw radii and use

'ø at centre' theorem.

We use 2 'previous' facts involving right øs

tangent ⊥ diameter . . . so, draw a diameter!

ø in semi-? = 90º . . . so, join RK!

O

N

Q

M

K

P

x

O

N

Q

M

K

P

R

x


PROVING THEOREMS

x y

2x

2y

x

y

x y

x

y

2x x

y 2y 2x 2y


radii

Note:

The bolded words in the statements are the approved 'reasons' to use.

? THEOREM PROOFS: A Visual presentation

Examinable

The LOGIC . . . The Situation Construction

1

chord

centre a perpendicular

line

radii 2

chord

centre midpoint of chord

The line drawn from the centre of a circle that bisects a chord is perpendicular to the chord. Theorem

Statement

The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord. Theorem

Statement



radius produced

exterior ∠s of Δs

x

x

y

y

2y 2x

3

equal base ∠s

x

x

y

y

radii 4

opposite ∠s in a


y

x

∠ at centre = 2 % ∠ at circumference

& ∠s about a point = 360º.

2y 2x

y

x

inscribed ∠

central ∠

arc

centre

The angle which an arc of a circle subtends at the centre is double the angle it subtends at any point on the circumference.

Theorem

Statement

The opposite angles of a cyclic quadrilateral are supplementary. Theorem

Statement

Copyright © The Answer Series ix

? THEOREM PROOFS: A Visual presentation, continued

5

Examinable


inscribed ∠

tangent

chord

This theorem is known

as the

Tan Chord theorem

Method 1:

radii

∠ at centre = 2 % ∠ at circumference

x

radius ⊥ tangent

x

90° – x

base ∠s of isosceles Δ

x

90° – x

sum of ∠s of Δ

x

2x

x

2x

x

The angle between a tangent to a circle and a chord drawn from the point of

contact is equal to the angle subtended by the chord in the alternate segment.

Theorem

Statement

Method 2:

diameter & join . . .

diameter ⊥ tangent

& ∠ in semi-?

x x x

x

90° – x

90° – x

x

∠s in the

same segment


Opposite ∠s : x + y = 180º Adjacent ∠s : x + y = 180º

∠s subtended

by an arc

(at the

circumference

of the circle)

∠ subtended by an arc (or chord)

(at the centre of the circle)

The LOGIC . . . The Situation

Theorem

Statements

The ∠ in a semi-?

is a right ∠.

∠s

subtended by

the same arc

are equal.

The exterior ∠ of a


= the interior opposite ∠.

FURTHER ? THEOREM PROOFS: A Visual presentation

diameter

y

x

180º

?

x

x

2x

2x

6

7

8

arc

Central angle is

a straight angle

Inscribed angle is

a right angle

∠s subtended by an arc (or chord)

(at the circumference)

Construction: radii

exterior angle


The LOGIC . . .

y

x

x

Copyright © The Answer Series xi

Method 1:

radii

Method 2:

chord and inscribed ∠

FURTHER ? THEOREM PROOFS: A Visual presentation, continued . . .

9

Again, the

bolded words

are the ‘approved

reasons' to use.

The LOGIC . . . The Situation

Theorem

Statement

Construction

x

radii ⊥ tangents congruent Δs

tan-chord theorem 4

x

x

x

sides opposite

equal base ∠s in Δ

x

x

x

Proofs to are not examinable, but,

the LOGIC is crucial when studying geometry.

6 9

Tangents to a circle

from a common point

are equal.


PROBLEM-SOLVING: An Active Approach

ACT!

Be Active . . .

Mark all the information on the drawing:

� equal or parallel sides � equal øs � right øs

� radii � diameter � tangents

Use all the Clues . . .

The information provides clues to facts:

� intersecting lines � adj. suppl. & vert. opp. øs ; øs about a point.

� parallel lines �� alternate, corresponding & co-interior øs

� equal radii or tangents (in triangles) �� equal base øs

� equal chords �� equal øs

� 90º angle �� the Theorem of Pythagoras, or . . .

� diameter � ø in semi-? = 90º ; diameter ⊥ tangent

� Δs � sum of the interior øs = 180º ; exterior ø of Δ ; equilateral Δs ;

isosceles Δs ; congruency, similarity, theorem of Pythagoras � s � sum of the interior øs of a quadrilateral = 360º ;

types & properties of all quadrilaterals

Recall the Theory systematically . . .

� Previous geometry

Revise previous geometry under the headings

angles � lines � triangles � quadrilaterals

� Circle geometry

Recall and apply the facts by group.

EXAMPLE 1

O is the centre of the circle

and diameter KL is produced

to meet NM produced at P.

ON || LM and ˆF = 76º. Calculate, giving reasons,

the sizes of :

(a) ˆ

1L (b) ˆ

1O

(c) ˆ

4M (d) ˆ ˆ

1 2N + N

(e) ˆ

1M (f) Prove that KG = GM

Answers

(a) ˆ

1L( )ˆ= F

= 76º � . . . øS in same segment

(b) ˆ

1O = corr. ˆ 1L . . . ON || LM

= 76º �

(c) ˆ

4M = ˆ

1

1

2O

= 38º � . . . ø at centre = 2 % ø at circumference

(d) ˆ

1N + ˆ

2N = 180º - 76º . . . opp. øs of c.q. are suppl.

= 104º �

(e) ˆKML = 90º . . . ø in semi-?

â ˆ 1M = 180º - (90º + 38º) . . . str. ˆNMP& ˆ

4M = 38º in 4.3

= 52º �

(f) ˆ

3G = corresponding ˆKML = 90º . . . ON || LM

i.e. line from centre, OG ⊥ chord KM â KG = GM �

F

K

N

L P

M

HO

76º

2 1

1 3

G

1

1

1

2

2

2

2 3

4

This is an excellent example testing so many

?-geometry theorems (excluding tangents).

F

K

N

LP

M

H O

76º

2 1

13

G

1

1

1

2 2

2

2 3

4

There are

3 theorems

in ? geometry

which involve

right angles.

A

C

T


EXAMPLE 2

(a) Complete the following by writing the appropriate missing word. If a chord of a circle subtends a right angle on the circumference,

then this chord is a . . . . . .

(b) A, B, C and D are points on the circle. BC produced and AD produced

meet at N. AB produced and DC produced

meet at M.

If ˆM = ˆN, prove that AC is a diameter of the circle.

Answers

(a) . . . diameter.

(b) Let ˆM = ˆN = x . . . given

& ˆ 1C = ˆ 2C = y . . . vert. opp. øs

Then:

ˆ

1B = x + y . . . ext. ø of ΔBMC

& ˆ 1D = x + y . . . ext. ø of ΔDCN

But, ˆ 1B + ˆ 1D = 180º . . . opp. øs of c.q.

â 2(x + y) = 180º

â x + y = 90º

i.e. ˆ 1B = ˆ 1D = 90º â AC is a diameter � . . . subtends a right ø

3 ways to prove that a quadrilateral is cyclic

We use the 3 converse statements for cyclic quadrilaterals. � Prove that one side subtends

equal angles at the two other points (on the same side).

� Prove that a pair of

opposite angles is supplementary.

� Prove that an exterior angle

of a quadrilateral is equal to the interior opposite angle.

2 ways to prove that a line is a tangent to a ?

We use the 2 converse theorem statements to prove that a line is a tangent to a ? if . . .

� . . . it is perpendicular to a radius

at a point where the radius meets

the circumference.

� . . . when drawn through

the end point of a chord,

it makes with the chord

an angle equal to an angle

in the alternate segment.

A

D

B

M

2 1

2

C

2 1

1

N

A

D

B

M

2 1

2

C

2 1

1

N

x

x

y

y

x + y

x + y

A: Be Active

C: Use all your Clues

T: Apply the Theory systematically, recalling

each group, and fact, one at a time.

ACT!

converse theorem

converse theorem

Prove that OP ⊥ AB;

then AB will be a tangentA

B P

O

Prove that

x = y;

then AB will be

a tangent

Note:

It is a good idea to draw a faint circle (as shown) to see

this converse theorem clearly.

A B

y

x

Either prove:

x1 = x2 or y1 = y2

or z1 = z2 or p1 = p2

Either prove:

Aˆ +Cˆ = 180º

or Bˆ +Dˆ = 180º

Either prove:

1ˆ = 2ˆ

or 3ˆ = 4ˆ

or 5ˆ = 6ˆ

or 7ˆ = 8ˆ

converse theorem

converse theorem

converse theorem

y1

y2 p2

p1

x1

x2

z1 z2

B

A

CD

1

2

3

4

5

6

7

8

Gr 12 Maths National November 2018: Paper 2

Q26 Copyright © The Answer Series: Photocopying of this material is illegal

2018

EUCLIDEAN GEOMETRY

Give reasons for your statements in QUESTIONS 8, 9 and 10

QUESTION 8

8.1 PON is a diameter of the circle centred at O.

TM is a tangent to the circle at M, a point on the circle.

R is another point on the circle such that OR || PM.

NR and MN are drawn. Let ˆ

1M = 66º.

Calculate, with reasons, the size of EACH of the following

angles :

8.1.1 ˆP 8.1.2 ˆ

2M (2)(2)

8.1.3 ˆ

1N 8.1.4 ˆ

2O (1)(2)

8.1.5 ˆ

2N (3)

8.2 In the diagram, ∆AGH is drawn. F and C are points on

AG and AH respectively such that AF = 20 units,

FG = 15 units and CH = 21 units. D is a point on FC

such that ABCD is a rectangle with AB also parallel to GH.

The diagonals of ABCD intersect at M, a point on AH.

8.2.1 Explain why FC || GH. (1) 8.2.2 Calculate, with reasons, the length of DM. (5)

[16]

QUESTION 9

9.1 In the diagram, JKLM is

a cyclic quadrilateral and

the circle has centre O. Prove the theorem

which states that

ˆ ˆJ + L = 180º. (5)

9.2 In the diagram, a smaller circle ABTS and a bigger circle

BDRT are given. BT is a common chord.

Straight lines STD and ATR are drawn.

Chords AS and DR are produced to meet in C, a point

outside the two circles.

BS and BD are drawn.

ˆA = x and ˆ

1R = y.

9.2.1 Name, giving a reason, another angle equal to :

(a) x (b) y (2)(2)

9.2.2 Prove that SCDB is a cyclic quadrilateral. (3)

9.2.3 It is further given that ˆ 2D = 30º and ˆAST = 100º.

Prove that SD is not a diameter of circle BDS. (4)

[16]

QUESTION 10

In the diagram, ABCD is a cyclic quadrilateral such that

AC CB and DC = CB. AD is produced to M such that

AM MC.

Let ˆB = x.

10.1 Prove that : 10.1.1 MC is a tangent to the circle at C. (5) 10.1.2 ∆ACB ||| ∆CMD (3)

10.2 Hence, or otherwise, prove that:

10.2.1 2

2

CM AM

ABDC

= (6)

10.2.2 AM

AB = sin2x (2)

[16]

TOTAL: 150

M

T

R

N

O

P

1

1

1 2

2

2 66º

F

B

20

D

A

15

C

21

H

M

G

M

K

J

O

L

D

A

R

C

B

1

1

2

2 3

2 x

3

S

T

4

1

2 2 y

1

1

D

A

C

B

1

1 2

x

M 1

2

2

Copyright The Answer Series 15

EXAMPLE 4

Make statements, with reasons,

1. In ?XPBA: about 1Pˆ and 1B

ˆ

2. In ?ABYQ: about 1Bˆ and AQY

ˆ

3. In quadrilateral APTQ: about 1Pˆ and AQT

ˆ

4. What can you conclude about quadrilateral APTQ?

Answers

1. In c.q. XPBA: 1Pˆ = 1B

ˆ . . . arc XA subtends øs in same segment

2. In c.q. ABYQ: 1Bˆ = AQY

ˆ . . . exterior ø of cyclic quad.

3. In quad. APTQ: 1Pˆ = AQT

ˆ . . . both = ˆ1B above

4. APTQ is a cyclic quad. . . . converse of exterior ø of cyclic quad.

EXAMPLE 5

Prove that PA

is a tangent to ?M.

Answers

In right-ød MAQ:

AM = 5 units . . . 3:4:5 ; Pythag.

â TM = 5 units . . . TM = AM = radii

â PM = 13 units

â PAM is a 5 : 12 : 13 !

i.e. PM2 = PA2 + AM2

â PAMˆ = 90º . . . converse of Pythag.

â PA is a tangent to ?M � . . . converse tan chord theorem

Mark the øs on the drawing as you proceed.

T

A

Q

P

1 2

2 1

X BY

3

1

2

1 2 1

2

Don't be put off by this drawing!

Direct your focus to one situation at a time

Q

P

M 12

8

3

4

T

A B

Q

P

M 12

8

3

4

T

A B

Module 9b: Circle Geometry

� Notes � Exercises � Full Solutions

See Gr 11 Maths 3 in 1

Study Guide


EXAMPLE 7

AB is a tangent to the circle at B and BD is a chord.

AD cuts the circle in E.

C is a point on BD so that ABCE is a cyclic quadrilateral.

AC, BE and CE are joined.

Prove that :

(a) AB = AC

(b) AC is a tangent to circle ECD at point C.

Answers

(a) Let 1Bˆ = x

then Dˆ = x . . . tan chord theorem

and 2Cˆ = x . . . øs in same segment

& Let 2Bˆ = y

then 2Aˆ = y . . . øs in same segment

3Cˆ = Dˆ + 2A

ˆ . . . ext. ø of Δ

= x + y

& ABCˆ = x + y . . . 1Bˆ = x and 2B

ˆ = y

AB = AC . . . sides opp = øs

(b) 2Cˆ =D

ˆ . . . both = x in (a) . . . converse tan chord thm

AC is a tangent to ?ECD at C.

1

DC

B

A

E 1

1

1

2

2

2

2

3

3

1

DC

B

A

E 1

1

1

2

2

2

2

3

3

y

y

x x

x


ENGLISH & AFRIKAANS

WWW.THEANSWER.CO.ZA

TAS

Proportionality, Similarity & The Theorem of Pythagoras



Proportionality, Similarity &

The Theorem of Pythagoras


WWW.THEANSWER.CO.ZA


PROPORTIONALITY, SIMILARITY &

THE THEOREM OF PYTHAGORAS

PROPORTIONALITY

� Ratio

Example 1: Determine the ratio BC : AB

Ratios can also be written as fractions:

BC

AB =

3

6 =

1

2 �

In Trigonometry, we have the sin, cos and tan ratios.

e.g. BC

AB = sin A

& sin A = 1

2 � ˆA = 30º

Example 2: Divide a 30 cm line in the ratio 2 : 3.

â AP = 12 cm and PB = 18 cm �

Note: AP

BP=2

3 � AP =

2

3PB, but,

AP

BA=2

5 � AP =

2

5AB

� Proportion

The Proportion theorem states:

If a line, PQ, is drawn parallel to one side of a triangle,

it divides the other two sides proportionally.

i.e. If PQ || BC, then a

b =

c

d.

The lengths of a, b, c and d, could be as follows:

a = 4 units ; b = 2 units ; c = 6 units ; d = 3 units

Then: a

b =

4

2 = 2 and

c

d =

6

3 = 2

â a

b =

c

d

Note: A proportion can be written in many ways:

If 4

2 =

6,

3 then

2

4 =

3

6 (invertendo) or

4

6 =

2

3 (alternando).

Both BC and AB

are lengths,

measured in

the same unit.

A

P Q

B C

4 6

3 2

These forms are all equivalent and imply that : 2 % 6 = 4 % 3.

Answer

The ratio BC : AB = 3 : 6 = 1 : 2 �

Alternatively,

let AP = 2k and PB = 3k;

then AB = 5k

â 5k = 30 cm

â k = 6 cm

A B 30 cm

P A B

3 parts 2 parts

A

P Q

B C

a c

d b

Answer

When two ratios are equal, e.g. ,

a c =

b d

we say that : a, b, c and d are in proportion, or,

that a and b are in the same proportion as c and d. We use Ratio to compare two quantities

of the same kind (in the same unit).

To summarise:

a c =

b d �

b d =

a c or

a b =

c d or ad = bc

A

CB

6 cm

3 cm

1

30º

60º

2 3

A

B C


THE PROPORTION THEOREM

The Theorem Statement:

i.e. PQ || BC � AP

PB=AQ

QC

The Converse Theorem Statement:

i.e. AP

PB =

AQ

QC � PQ || BC

Worked Example 1 Worked Example 2

Find x : Find x :

Answer Answer

Because of the || lines, and the Because of the || lines, and the proportion theorem above, we know proportion theorem above, we know that

that the ratio x : 2 will equal the ratio 6 : 3 the ratio x : 5 will equal the ratio 3a : a (= 3 : 1) â By inspection: x = 4 units â By inspection: x = 15 units

Worked Example 3

In the figure, RT | | AB, RS | | AC and DR

RA =

2

5.

3.1 Write down the values of the following ratios:

(a) DT

TB = (b)

DS

DC =

3.2 Prove that TS || BC

Answers

3.1 (a) DT

TB =

DR

RA =

2

5 � . . .

(b) DS

CD =

DR

AD =

2

7 � . . .

3.2 In ΔDBC:

DT

TB =

DS

SC . . . both =

DR

RA =

2

5

â TS || BC � . . . converse of proportion theorem

A

B

C

S

D

R

T

A

P Q

B C

or, by calculation: 2

x = 6

3

% 2) â x = 2 % 2

â x = 4 units �

or, by calculation: 5

x = 3a

a

% 5) â x = 5 % 3

â x = 15 units �

In ΔDBA:

RT || AB ; prop. theorem

In ΔDCA:

RS || AC ; prop. theorem

A

P Q

B C

x 6

2 3 x

a 5

3a

A line parallel to one side of a triangle

divides the other two sides proportionally.

If a line divides two sides of a triangle

proportionally, the line is parallel to the third side.

Note: When applying these statements, focus on one triangle

at a time, and apply either one fact or the other.

Remember:

AB : BC = 2 : 5, while AB : AC = 2 : 7

i.e. AB

AC =

2

7

â AB = 2

7AC

A

B

C

Question

Solution


Question

Solution

Worked Example 5

In PQR the lengths of PS, SQ, PT and TR are

3, 9, 2 and 6 units respectively.

5.1 Give a reason why ST || QR.

5.2 If AB || QP and RA : AQ = 1 : 3,

calculate the length of TB.

Answers

5.1 In PQR: PS

SQ =

3

9 =

1

3 &

PT

TR =

2

6 =

1

3

PS

SQ =

PT

TR

â ST || QR � . . . converse of proportion thm

5.2 In RPQ: RB

RP =

RA

RQ =

1

4 . . . proportion theorem ; AB || QP

â RB = 1

4RP

= 2 units . . . RP = PT + TR = 8 units

â TB = 4 units �

P

S T

B

RA Q

P

S T

B

2 3

96

3 parts 1 partRA Q

RA:AQ = 1:3


� Δs on the same base

and between the same || lines

have equal areas.

ΔABC = ΔDBC in area

� When Δs have the same height,

the ratio of their areas equals

the ratio of their bases.

Δ

Δ

Area of ABC

Area of ACD =

x

1

2

1

2

.h

y.h =

y

x

Proving the Proportion theorem

Be sure to revise the following two concepts involving areas of triangles.

These concepts are used in the proof of the proportion theorem which follows.

IMPORTANT CONCEPTS REQUIRED

THE PROOF OF THE PROPORTION THEOREM

Given: ΔABC with DE || BC, D & E on AB & AC respectively.

To prove: AD

DB =

AE

EC

Construction: Join DC & BE

Proof: Δ

Δ

Area of ADE

Area of DBE =

1

2

1

2

AD.h

DB.h

= AD

DB

Similarly: Δ

Δ

Area of ADE

Area of EDC =

AE

EC

1

2

1

2

AE.h

.hEC

′

′

But: ΔDBE = ΔEDC . . . on the same base DE ; between || lines, DE & BC

and: ΔADE is common

â Δ

Δ

Area of ADE

Area of DBE =

Δ

Δ

Area of ADE

Area of EDC

â AD

DB =

AE

EC �

E

B C

A

D

h′ h

B

A

DC

A

B DC

h

x y

These Δs have the same base, BC,

and the same height (since they lie

between the same || lines).

These s have a common vertex, A,

and therefore the same height.


= 1

2

1

2

ah

bh =

a

b =

1

2

1

2

ch

dh

'

'

= c

d But: =

a

b =

c

d

These 2 Δs have

equal areas

(same base &

same height H)

These 2 Δs have

common height h

These 2 Δs have

common height h'

PROPORTION THEOREM PROOF: A Visual presentation

The Theorem Statement:

A line drawn parallel to one side of a triangle

divides the other two sides proportionally.

The Construction The Situation Heights of Δs

d

c

h' BaseBASE

H

a

b

c

d

Create 2 Δs

h

h'

base a

height (H)

base c

a

b

h

Bases

Parallel lines in a Δ

a

b

c

d

The

same


EXERCISE 1

PROPORTION THEOREM AND APPLICATIONS

Answers on page 17

1. In the figure A and B are points on PQ and PR

such that AB || QR. AR and BQ are drawn. Answer the following questions, which refer to

a theorem. You need to redraw the sketch.

1.1 Complete: Δ

Δ

area APB

area AQB =

. . . .

. . . .

1.2 Complete: Δ

Δ

area APB

area ABR =

. . . .

. . . .

1.3 What can you say about the area of ΔAQB and the area of ΔABR, and why? 1.4 What can you deduce from 1.1, 1.2 and 1.3? 1.5 Give the wording of the theorem which is under consideration here. 2.1 Complete the following theorem by writing down the missing word(s) only.

"A line parallel to one side of a triangle divides the two other sides . . ."

2.2 In the accompanying figure, MS || QR. Furthermore, PM = x cm,

MQ = 2 cm, PS = (x + 2) cm

and SR = 3 cm. Determine, without giving reasons,

the value of x.

2.3 The following measurements

are known in the given sketch: AD = 3x - 1

BD = 7x - 6

AE = 3

CE = x + 2

Determine the integral value(s) of x for which DE || BC.

2.4

2.5 In the accompanying figure, BF || CG || DH

and DG || EH. Furthermore, AB = 1 cm, BC = 3 cm and CD = 1 cm.

2.5.1 Write down the values of the following ratios:

(a) AG

GH (b)

AD

DE

2.5.2 Determine the length of DE.

3.1 Complete the following theorem statement:

If a line is parallel to one side of a triangle, then the line divides the other

two sides in proportion, and conversely if . . . . . . . , then . . . . . . .

3.2 Quadrilateral WXYZ is given. R is a point on WZ and T is a point

on YZ such that RT is parallel to

diagonal WY.

If S is a point on diagonal XZ such that

SR || XW, prove that ST || XY.

P

M S

Q R

x + 2 x

32

A

D

B

E

C

A

D

B

E

C

3 cm

5 cm x cmIn ΔABC, DE || BC and AB = 6 cm.

AD = x cm, AE = 5 cm and EC = 3 cm.

Determine the length of AD,

i.e. find x.

A

B

F

C

D

E

G

H

1

3

1

W R

Z

T

Y X

S

P

A B

Q R


4. In the diagram below HJKL is a parallelogram, with the diagonals

intersecting at M.

ˆJHK = 90º. JK is produced to S. N is a point on HL. NS intersects JL at F.

HJ = 6 units ; HK = 8 units ; KS = 5 units ; FL = 13 units

4.1 Determine, with reasons, the following ratios in simplified form: (a) JK : KS (b) JM : MF

4.2 Hence, prove that HK || NS

5. In the accompanying figure, DF || BC

and AF

FE=FC

EB.

Prove that ADEF is a trapezium.

SIMILARITY

� The definition of similarity

The conditions for polygons to be similar are:

A : the polygons must be equiangular, AND

B : their corresponding sides must be in proportion.

We will show that, for triangles :

i.e. If, in ΔABC & ΔPQR,

ˆA = ˆP , ˆB = ˆQ and ˆC = ˆR ,

then AB

PQ =

BC

QR = .

AC

PR

â ΔABC ||| ΔPQR

and, conversely . . .

If, in ΔABC & ΔPQR,

AB

PQ =

BC

QR =

AC

PR, then

ˆA = ˆP , ˆB = ˆQ and ˆC = ˆR.

â ΔABC ||| ΔPQR

A

B C

P

Q R

A

x B C

P

x Q R

H N L

J K S

M

F

A

F

C

B

E

D

If B holds, then A holds â similar

A: Be Active


T: Apply the Theory systematically

ACT!

If A holds, then B holds â similar

When polygons are similar, they have the same shape, but NOT necessarily

the same size. One figure is an enlargement or reduction of the other.


A

B

E

C

D

10

84

Worked Example 1

In the figure, BA || ED, AB = 8, AC = 4 and DE = 10 units. 1.1 Name a pair of similar triangles (in the correct order).

Explain why they are similar. 1.2 Calculate the length of EC.

Answers

1.1 In Δs ACB and ECD . . .

ˆ Â = E . . . alternate øs; BA || ED

ˆ ˆB = D . . . alternate øs; BA || ED

Also, ÂCB = ÊCD . . . vert. opposite øs

â ΔACB ||| ΔECD � . . . AAA

1.2 ∴ EC ?

AC =

ED

AB

=

CD

CB . . . ACB ||| ECD

â AC

EC =

ED

AB . . .

â 4

EC =

10

8

% 4) â EC = 5 units �

Worked Example 2

Say, with reasons, whether the following pairs of triangles are similar or not.

2.1

2.2 2.3

Answers

2.1 ΔLMN ||| ΔEFH � because: 51

17 equals 3;

69

23 equals 3 and

75

25 equals 3

∴ LM

EF =

MN

FH =

LN

EH, i.e. the sides are in proportion.

2.2 ΔPQR ||| ΔABC � because: while PQ

AB =

12

6 = 2 and

PR

AC =

18

9 = 2 ,

QR

BC =

9

3 = 3, which is ≠ 2

â The sides are not in proportion.

2.3 ΔBAC ||| ΔEDC; because, in these triangles:

ˆB = Ê . . . both = x

Â = ˆD . . . both = y

[â 3rd ø equal too! ]

â ΔBAC ||| ΔEDC � . . . AAA

E A

B C D

y

y

x

x

N

69

L

M 75

51

H

2317

25

E

F

Note: These 2 Δs are not congruent.

Even though there are 2 øs

and a side equal, the sides do

not correspond.

When creating your equation, place what

you are looking for, EC in this case, on

the top LHS. It eases the calculation.

Note: As soon as you have shown Δ's to be similar by showing they are

equiangular, you can claim that their sides are in proportion

and write this down without having to look back at the diagram.

i.e. if Δ�ACB ||| Δ�ECD , then: AC

EC=AB

ED=CB

CD or :

NB: The triangles must be named in the correct order!

It is then useful to mark off what you are looking for with a ?

(or its letter if it has a letter) and a for the sides you have lengths for.

the same sides,

but inverted.

A

9 6

3BC

P

R9 Q

1218

It is essential when naming the Δs

,

to order the letters according

to the equal øs

, especially for

writing out the proportional

sides which follow.


In the same figure above,

ΔABC can be seen as

an enlargement of ΔADE

and the sides of these

triangles are proportional.

â y

30 =

12

12 + 8

15or ⎛ ⎞⎜ ⎟⎝ ⎠15 + 10

Worked Example 3

Find the values of x and y in the

figure alongside.

Answer

� In ΔABC: 10

x =

8

5 . . . DE || BC; proportion theorem

% 10) ∴ x = 1

46 cm �

� ∆ADE ||| ∆ABC � 12

y =

8

5 . . .

BC

DE =

AC

AE; proportional sides

% 12) ∴ y = 1

27 cm �

Similar Δs vs.

Proportion Theorem Application

Worked Example 4

Find x and y in the sketch alongside

� The Proportion theorem (finding x)

In ΔABC: DE || BC

15

x =

8

12

% 15) â x = 10 units

� Similar triangles theorem (finding y)

In Δs ADE and ABC:

(1) Â is common

(2) ÂDE = ˆB . . . corresponding øs; DE || BC

[& ÂED = ˆC . . . corresponding øs; DE || BC ]

â ΔADE ||| ΔABC . . . equiangular Δs

â DE

BC =

AD

BA or

AE

CA . . .

â y

30 =

12

20

% 30) â y = 18 units

The proportion theorem

does NOT refer to the

lengths of the parallel lines,

only to AB and AC

and their segments.

Note:

Distinguish between the applications of

the similar ∆s and proportion theorems!

(See next column.) It is only by using the similarity of the triangles,

that we can relate the lengths of the parallel sides to

the lengths of the other 2 sides of the triangles.

Note: The Theorem of Pythagoras can be proved using

similar ∆s (see page 13 & 14). Consider this proof an ideal

'worked example' of the application of similar triangles.

The unknown

�

Note: DE

BC =

AD

DB or

AE

EC

because BC is a side of ΔABC,

while DB and EC are not.

A

D E

B C

12

8

15

x

30

y

A

D E

B C

x 5 cm

12 cm

y

3 cm

10 cm


Stage 2:

corresponding øs

Stage 3:

parallel lines

Stage 1:

congruency

Stage 4:

proportions

SIMILAR ΔS: STATEMENTS & PROOF

STATEMENT: Equiangular Δs � proportional sides � SIMILAR Δs

PROOF:

Given: ABC & DEF with Â = ˆD, ˆB = Ê & ˆC = ˆF

Required to prove: AB

DE =

AC

DF =

BC

EF

Construction: Mark P & Q on DE & DF such that DP = AB & DQ = AC Proof: In s DPQ & ABC

(1) DP = AB . . . construction

(2) DQ = AC . . . construction

(3) ˆD = Â . . . given

â DPQ ≡ ABC . . . SøS

â ˆ1

P = ˆB

= Ê . . . given

But DP = AB and DQ = AC . . . construction

â AB

DE =

AC

DF

Similarly, by marking P and R on DE and EF such that

PE = AB and ER = BC, it can be proved that: ABDE

= BCEF

â ABDE

= ACDF

= BCEF

�

â ABC and DEF are similar

â PQ || EF . . . corresponding øs equal

â DP

DE =

DQ

DF . . . (prop thm; PQ || EF)

The

focal

point

A

CB

D

FE

P Q1

If two triangles are

equiangular, then their sides

are proportional and, therefore,

they are similar.


NOT listed as

examinable in the 2021

Exam Guidelines.

Stage 4:

return to the start !

(∆3 ! ∆2 ! ∆1)

Stage 2:

proportions

compared

Stage 1:

equiangular ∆s

(∆1 & ∆2)

Stage 3:

congruency

(∆2 & ∆3)

THE CONVERSE: Proportional sides � equiangular Δs � SIMILAR Δs

PROOF (Optional):

Given: s ABC & DEF with DE

AB =

BC

EF =

DF

AC

Required to prove: ABC & DEF are equiangular

Construction: ˆGEF ˆ

1( )x = ˆB & ˆGFE ˆ

1(y ) = ˆC G & D on opposite sides of EF

Proof: In s ABC & GEF

(1) ˆ

1x = ˆB . . . construction

(2) ˆ

1y = ˆC . . . construction

â ABC & GEF are equiangular

â GE

AB =

BC

EF =

GF

AC . . .

â In s DEF & GEF

(1) GE = DE . . . proved above

(2) GF = DF . . . proved above

(3) EF is common

â DEF ≡ GEF . . . SSS

â ˆ2

x = ˆ1x & ˆ

2y = ˆ

1y

= ˆB = ˆC . . . construction

â also ˆD = ˆA . . . 3rd ø of the two s i.e. s ABC & DEF are equiangular . . . (∆1 & ∆3)

â ABC and DEF are similar . . . as for 1st proof!

proportional sides of equiangular s

(applying the 'original' theorem)

D

F E

A

CB y1

y2

x1

x2

G

1 3

2

Comparing this with the 'given' :

â GE

AB =

DE

AB and

GF

AC =

DF

AC . . .

BCall =

EF

â GE = DE and GF = DF

The

focal

point

Triangles with sides in proportion, are equiangular and,

therefore, they are similar.

Copyright © The Answer Series: Photocopying of this material is illegal 13

2 Equiangular Δs : ΔABC & ΔDEF

The 2 shaded Δs are congruent (SS)

ˆ1 = ˆ2

But ˆ2 = ˆ3 . . . given

ˆ1 = corresponding ˆ3

The horizontal lines are parallel

You actually mark the lengths of

the small Δ onto the sides of the big Δ.

Then join their endpoints.

SIMILAR ΔS THEOREM PROOF: Visualised


Imagine copying the

small Δ on the big Δ

The Statement:

Equiangular Δs are similar

A

B C

The sides of this Δ are proportional.

i.e. p

x = y

q . . . Proportion theorem

But: AB = x & AC = y

AB

DE = AC

DF

Similarly, it can be proved that:

AB

DE = BC

EF

AB

DE = BC

EF = AC

DF

ΔABC ||| ΔDEF

y x

A

B C

y x

D

E F

y x

p q

D

E F

3

1

2


EXAMPLE 3 (National November 2017 P2, Q10)

In the diagram, W is a point on the

circle with centre O. V is a point on OW. Chord MN is drawn such that

MV = VN. The tangent at W meets

OM produced at T and

ON produced at S. (a) Give a reason why OV ⊥ MN. (b) Prove that:

(i) MN || TS

(ii) TMNS is a cyclic quadrilateral

(iii) OS .MN = 2ON .WS (a Grade 12 question)

Answers

(a) Line (OV) from centre to midpoint of chord (MN) �

(b) (i) ÔWS = 90º . . . tangent ⊥ radius

â ÔVN = ÔWS (= 90º)

â MN || TS . . . corresp. øs equal (ii)

ˆ

1M = ˆ

1N . . . øs opposite equal radii

= ˆS . . . corresp. øs; MN || TS

â TMNS is a cyclic quadrilateral . . .

The most 'basic'

way to prove

lines || is: alt. or corresp.

øs equal or

co-int. øs suppl.

converse ext. ø

of cyclic quad.

Shade the

quadrilateral

TMNS

O

T

M

V

W

S

N

1 1

1

2 2

2

In this case, the midpoint of the chord is given, and

we can conclude that OV ⊥ MN because of that.

Note: Analyse the information and the diagram.

So far, we have used and applied a 'centre' theorem, in (a).

Another clue is the ' tangent' at W.

Think about tangent facts . . . .

We chose � and proved

that the exterior ø of

quadrilateral TMNS

= the interior opposite ø

There are 3 ways to prove that a quadrilateral is

a cyclic quadrilateral – choose 1:

� � �

Prove that: Prove that: Prove that:

x + y = 180º Ext. ø = int. opp. ø A side subtends equal øs

at 2 other vertices

x

y

O

T

M

V

W

S

N

1 1

1

2

2

2


(iii)

In s OVN and OWS

ˆ 2O is common

ˆOVN = ˆOWS . . . corresp. øs ; MN || TS

â OVN ||| OWS . . . øøø

â OS

ON =

WS

VN= OW

OV

. . . equiangular s

â OS .VN = ON .WS

But VN = 1

2MN . . . V midpoint MN

â OS .1

2MN = ON .WS

% 2)â OS .MN = 2ON.WS �

Let's 'arrange' the sides to suit the question.

This question looks like ratio and proportion.

Mark the sides on the diagram. The sides appear to involve OWS, which has VN || WS,

(even though MN = 2VN) . . . Maybe apply the proportion theorem in this ? But, the sides in the question involve the horizontal sides

WS and VN.

So, proportion theorem is excluded.

We will use similar s !

Copyright © The Answer Series: Photocopying of this material is illegal 14

THE THEOREM OF PYTHAGORAS

� THE STATEMENT

In ΔABC: ˆC = 90º � c2 = a2 + b2

� THE CONVERSE STATEMENT

In ΔABC: c2 = a2 + b2 � ˆC = 90º

Given: ΔABC with ÂBC = 90º

Required to prove: AC2 = AB2 + BC2

Construction: Draw BD AC

PROOF:

Let Â = x ; then ˆ 1B = 90º – x . . . ø sum of ABD

â ˆ 2B = x

â ˆC = 90º – x . . . ø sum of BCD

: In Δs ABD and ACB:

(1) Â is common

(2) ˆ

1B = ˆC . . . = 90º – x

[& ÂDB = ÂBC = 90º]

â ABD ||| ACB . . . AAA

â AC

AB =

AD

AB

BD=

BC

â AB2 = AC . AD

: Similarly, by proving BCD ||| ACB:

â BC2 = AC . DC

: â AB2 + BC2 = AC .AD + AC .DC

= AC(AD + DC)

= AC . AC

= AC2

A

B C

c b

a

A

B C

c b

a

BCD ||| ACB

� AC

BC =

CD

BC

â BC2 = AC .CD

Here, BC is the common side :

BC2 = . . .

A

B C

D

In a right-angled Δ, the square on the hypotenuse

equals the sum of the squares on the other two sides.

If the square on one side of a triangle equals the

sum of the squares on the other two sides,

then the angle between these two sides is a right angle.

AB is the common side

of the 2 triangles:

AB2 = . . .

A

B C

D

A

B C

D

1

2

Proving the Theorem of Pythagoras, using similar triangles


NOT listed as examinable in the

2021 Exam Guidelines.

The Theorem of Pythagoras: The proof in stages

Similarly:

The Situation

A

B C

c

a

b

Construction

D

B

The Theorem Statement

In a right-angled Δ,

the square on the hypotenuse

equals the sum of the squares

on the other two sides.

The Method

The Conclusion

B C

b

A

c

D

In ΔAB D & ΔAC B : Common ˆA & a right each

ΔABD || | ΔACB

AC

AB =

AD

AB

AB2 = AC. AD

i.e. c2 = AC. AD . . .

A

B C

c

a

b

D

In ΔCB D & ΔCA B : Common ˆC & a right each

ΔCBD || | ΔCAB

CB

CA =

CD

CB

CB2 = CA.CD

i.e. a2 = AC.CD . . .

& :

c2 + a2 = AC . AD + AC . CD

= AC(AD + CD)

= b.b . . . AC = b

= b2

B C

b

A

c

D

a

16 Copyright © The Answer Series: Photocopying of this material is illegal

Similar s : Advice for problem-solving

� The symbol for similar Δs is: |||

& the symbol for congruent Δs is:

� If we write ΔABC ||| ΔDEF, with the letters in this order, it means

that ˆA = ˆD , ˆB = ˆE and ˆC= ˆF and the sides will also correspond . . .

AB

DE = BC

EF = AC

DF . . .

This is very important because it means we don't have to read off

the sides from the drawing, provided we have the letters in

the correct order !

� It is usually a good idea to write down all three ratios,

maybe bracketing the one not immediately required

– it may well be needed later in the question.

� We can prove that triangles are similar

either by:

� proving them equiangular (2 øs are sufficient) . . . The Theorem

or, by:

� proving that their sides are proportional. . . . The Converse Theorem

� When two pairs of angles of two triangles have been

proved equal, mark the third angles as being equal too.

It could be required later in the question.

� When asked to prove a proportion, e.g. AB

DE= BC

,EF

mark the 4 sides (AB, DE, BC & EF) on the figure.

It will then be clear which triangles need to be proved similar.

If asked to prove PQ2 = PR.PS, we need to mark PQ twice,

because it would be a common side of the two triangles

– see the proof of the Theorem of Pythagoras illustrating this.

� Always remember to accumulate facts as you work through

a sum, i.e. mark the facts on the figure as you prove them.

� Distinguish between the applications of the

proportion and the similar Δs theorems:

Given: AQ : QC = 5 :2 and PQ || BC

THE PROPORTION THEOREM

AP :PB = 5 :2

Also: AP : AB = AQ : AC = 5 :7 & PB : AB = QC : AC = 2 :7

whereas

. . . because PQ and BC are the sides of similar Δs APQ & ABC

PQ : BC = 5 : 7 SIMILAR TRIANGLES THEOREM

|||

�ABC

�DEF

A

P Q

B C

2 parts

5 parts

Be sure to know and apply the

theorem statements accurately !


EXERCISE 2: SIMILAR TRIANGLES

Answers on page 18

1.1 1.2

2. Make a neat copy of this sketch and fill in

all the other angles in terms of x.

Reasons are not required. 2.1 Complete the following statement:

∆ABE ||| ∆ . . . ||| ∆ . . . 2.2 If BC = 18 cm and BE = 12 cm,

calculate the length of 2.2.1 AE 2.2.2 AB correct to two decimals.

2.3 Hence calculate the area of rectangle ABCD to the nearest cm2.

3. In the figure, P is a point on SQ such

that ˆSTP = ˆTQS.

TS = 51 mm, PS = 32,6 mm and TP = 29 mm.

3.1 Prove that STP ||| SQT. 3.2 Calculate QT (answer correct to one decimal place).

4. In the figure, ˆCDB = ˆABC , AD = 5 and DC = 4. 4.1 Complete: CBA ||| . . . . . 4.2 Calculate the length of BC.

(Show all working details.)

5. Find, with reasons, the lengths of

x and y.

6.1 6.2

Prove: BA .BD = BC .BE Prove: AB2 = AC.AD

7. The accompanying figure shows ABC with DE || AC.

AC = 14, DE = 6 and EC = 4 units. 7.1 BDE ||| . . .

7.2 Complete: BE

. . . =

DE

. . .

7.3 Hence calculate the length of BE.

7.4 Determine the value of BD

AD.

8.1 In the figure alongside, prove that GDF ||| GED.

8.2 ˆE = . . . . . 9. Complete with reasons:

9.1 a

b = . . . . .

9.2 c

d = . . . . .

A E D

CB

x

B A

C4

5

D

T

S

3x

P

Q R

5 4

2

y

b d

c a

3

7 FE G

4,56

D

3

5 4

4

A

D

B

C

E

6

14

A

CB

P

Q R

ABC ||| . . . .

A

CB

D

E F

3236

28 27

24 21

CAB ||| . . . .

P Q S

T

32,6 mm

51 mm29 mm

E

A

D

B C

A

D

B

C

For more examples, see the

Topic Guide at the start of Section 2

in The Answer Series

Gr 12 Maths 2 in 1


ANSWERS TO EXERCISES

EXERCISE 1: PROPORTION THEOREM

AND APPLICATIONS

Questions on page 7

1.1 AP

AQ �

1.2 PB

BR �

1.3 They’re equal because they lie on the same base (AB)

and between the same || lines (AB & QR). �

1.4 AP PB =

AQ BR �

1.5 Bookwork

2.1 ". . . proportionally"

2.2 x

2 = x + 2

3 . . . proportion theorem

â 3x = 2x + 4

â x = 4 �

2.3 DE || BC � x

x

3 - 1

7 - 6 =

x

3

+ 2 . . . proportion thm.

â (3x - 1)(x + 2) = 3(7x - 6)

â 3x2 + 5x - 2 = 21x - 18

â 3x2 - 16x + 16 = 0

â (3x - 4)(x - 4) = 0

â x = 4

3 or 4

Check x = 4: LHS = x

x

3 - 1

7 - 6 = 11

22 = 1

2

& RHS = x

3

+ 2 = 3

6 = 1

2 Solution: x = 4

2.4 x

6 = 5

8 . . . proportion thm.; DE || BC

â 8x = 30

â x = 30

8

= 33

4

2.5.1 (a) AG

GH = 4 � . . . prop thm; CG || DH

(b) AD

DE = 4 � . . . prop thm; DG || EH

2.5.2 DE

AD = GH

AG

â DE

5 = x

x4 = 1

4

% 5) â DE = 5

4 �

3.1 Theorems

3.2 In ZWY:

ZT

TY = ZR

RW . . . RT || WY; proportion theorem

which, in ZWX,

= ZSSX

. . . RS || WX;

proportion thm.

â In ZXY we have

ZTTY

= ZSSX

â ST || XY � . . . converse of proportion theorem

A

B

F

C

D

E

G

H

1

3

1 x

3x

x

Note: 'integral' value(s) required

x 43

W RZ

T

Y X

S

P

A B

Q R

P

M S

Q R

x + 2 x

32

A

D

B

E

C

A

D

B

E

C

3 cm

5 cmx cm


4.1

(a) In JHK:

JK = 10 units . . .

JK:KS = 10:5 = 2 :1

(b) In JHM:

HM = 4 units . . . diagonals of a parallelogram

JM2 = 62 + 42 . . . ˆJHM = 90º; Pythagoras

= 52 JM = 52 = 4 13 = 2 13

ML = 2 13 . . . diagonals of ||m bisect one another

MF = 13 . . . FL = 13

JM:MF = 2 13 : 13 = 2 :1

4.2 In JFS: JK:KS = JM:MF . . . in (a) & (b)

MK || FS . . . converse of proportion theorem

HK || NS

5.

AF

FE = FC

EB � AF

FC = FE

EB . . . a c a b

= =b d c d

�

But, in ABC: AF

FC = AD

DB . . . prop. thm.; DF || BC

â FE

EB = AD

DB

â In ABF, DE || AF . . . converse of prop. thm. â ADEF is a trapezium � . . . 1 pair opp. sides ||

Now, try some

CHALLENGING QUESTIONS . . .

EXERCISE 2: SIMILAR TRIANGLES

Questions on page 16

1.1 ABC ||| RPQ� � � �

1.2 CAB ||| FED . . . sides in ratio 8 : 7 : 9

2.

2.1 ABE ||| ECB ||| DEC 2.2.1 ABE ||| ECB

� AE

BE = BE

BC . . . sides in proportion

% BE) â AE = 2

BE

BC =

212

18 = 8 cm

2.2.2 AB2 = 122 - 82 = 80 . . . Thm. of Pythagoras

â AB = 80

j 8,94 cm 2.3 Area of rectangle ABCD = BC % AB . . .

= 18 % 8,94

j 161 cm2

A E D

C B

x

90º- x

12

x

90º-x

90º- x

18x

length %

breadth

NB: The order of the letters

must correspond

with the equal angles

H N L

J K S

M

F

ˆJHK = 90º; Pythagoras

3:4:5 = 6:8:10

A

F

C

B

E

D

A: Be Active


T: Apply the Theory systematically

ACT!

See Section 3

Page 255 in

The Answer Series

Gr 12 Maths 2 in 1


3.1 In s STP and SQT

(1) ˆS is common

(2) ˆSTP = ˆQ . . . given

â STP ||| SQT . . . equiangular Δs

3.2 â QT

TP = ST

SP . . . sides in proportion

â QT

29 = 51

32,6

% 29) â QT = 51 29

32,6

%

j 45,4 mm 4.1 CBA ||| CDB . . .

4.2 â BC

CD = AC

BC

â BC2 = CD.AC . . .

= 4.9 = 36

â BC = 6 units 5. In PQR:

x

3 = 4

2 . . .

â x = 6 units

QRP ||| QTS . . . equiør ∆s . . .

â PR

ST = PQ

QS

â y

5 = 6

4

% 5) â y = 7 1

2units

6.1

In s BCA and BDE

(1) ˆB is common

(2) ˆC = ˆBDE . . . given

â BCA ||| BDE . . . equiangular Δs

â BA

BE = BC

BD

â BA.BD = BC.BE 6.2

In s ABD and ACB

(1) Â is common

(2) ÂBD = ˆC . . . given

â ABD ||| ACB . . . equiangular Δs

â ABAC

= ADAB

â AB2 = AC.AD

7.1 BDE ||| BAC . . . equiør ∆s . . .

7.2 â BE

BC = DE

AC . . . prop. sides

7.3 â BE

BE + 4 = 6

14 . . . sides of sim. ∆s

â 14BE = 6BE + 24

â 8BE = 24

â BE = 3 units

7.4 BD

AD = BE

EC = 3

4 . . . DE || AC ; prop. theorem

8.1

GD

GE = 6

9 = 2

3;

DF

ED = 3

4,5 = 6

9 = 2

3; and

GF

GD = 4

6 = 2

3

â GDF ||| GED . . . proportional sides

8.2 Ê = ˆGDF . . . øs in similar ∆s in 8.1

9.1 a

b = 7

3 . . . || lines ; proportion theorem !

9.2 c

d = 7

10 . . . proportional sides of similar s !

Now, try some

CHALLENGING QUESTIONS . . .

P Q S

T

32,6 mm

51 mm 29 mm

common ø &

corresponding øs

The order of the letters

must be correct !

proportional

sides

T

S

3 x

P

Q R

54

2

y

We need s BCA & BDE A

B C E

D

We need s ABD & ACB

Note: Side AB is common to the two s

D

B

C

A

NB: BDAD

DEAC

B A

C4

5

D

A

D

B

C E

6

14

4

F E G

4,5 6

D

5 4

3

Note:

It is a good idea to

place the required

length (QT) in the

top left position of

the proportion.

See Section 3

Page 256 to 262 in

The Answer Series

Gr 12 Maths 2 in 1

ST || PR ;

prop. thm.

common ø &

corresponding øs


See The Answer Series

Gr 12 Maths 2 in 1

Study Guide

For more practice, see the TOPIC GUIDE (on Page 148)

at the start of SECTION 2: The Exam Paper 2s

and

The Challenging Questions (Pp 255 – 262)

in the SECTION 3 NEW

PLEASE NOTE

These Geometry materials (Booklets 1 to 4) were created and

produced by The Answer Series Educational Publishers (Pty) (Ltd)

to support the teaching and learning of Geometry in high schools

in South Africa.






ENGLISH & AFRIKAANS

WWW.THEANSWER.CO.ZA

TAS


2021

9. In the diagram, PQRS is a cyclic

quadrilateral. PS is produced to W.

TR and TS are tangents to the circle at

R and S respectively.

Tˆ = 78º and Qˆ = 93º.

9.1 Give a reason why ST = TR. (1)

9.2 Calculate, giving reasons, the size of:

9.2.1 2Sˆ 9.2.2 3Sˆ (2)(2) [5]

P

R

S

T

1

1

2

2

3

78º

93º

Q

W


2021

9. In the diagram, PQRS is a cyclic

quadrilateral. PS is produced to W.

TR and TS are tangents to the circle at

R and S respectively.

Tˆ = 78º and Qˆ = 93º.

9.1 Give a reason why ST = TR. (1)

9.2 Calculate, giving reasons, the size of:

9.2.1 2Sˆ 9.2.2 3Sˆ (2)(2) [5]

MEMOS

9.1 Tangents from a common point.

9.2.1 2Sˆ =

2Rˆ . . . øs opposite equal sides

= 1

2(180º – 78º) . . . ø sum of Δ

= 51º �

9.2.2 3Sˆ + 2Sˆ = Qˆ . . . ext. ø of cyc. quad.

∴ 3Sˆ + 51º = 93º

∴ ˆ 3S = 42º �

P

R

S

T

1

1

2

2

3

78º

93º

Q

W


2021

10. In the diagram, BE and CD are diameters

of a circle having M as centre. Chord AE

is drawn to cut CD at F. AE CD.

Let Cˆ = x.

10.1 Give a reason why AF = FE. (1)

10.2 Determine, giving reasons, the size

of 1Mˆ in terms of x. (3)

10.3 Prove, giving reasons, that AD is

a tangent to the circle passing

through A, C and F. (4)

10.4 Given that CF = 6 units and

AB = 24 units, calculate, giving

reasons, the length of AE. (5) [13]

C E

F

A

B D

M

1

1

2

2

3

3

x


2021

10. In the diagram, BE and CD are diameters

of a circle having M as centre. Chord AE

is drawn to cut CD at F. AE ⊥ CD.

Let Cˆ = x.

10.1 Give a reason why AF = FE. (1)

10.2 Determine, giving reasons, the size

of 1Mˆ in terms of x. (3)

10.3 Prove, giving reasons, that AD is a

tangent to the circle passing through

A, C and F. (4)

10.4 Given that CF = 6 units and

AB = 24 units, calculate, giving

reasons, the length of AE. (5)

[13]

MEMOS

10.1 MF ⊥ AE, i.e. line from centre ⊥ to chord �

10.2 1

Mˆ = 1

2Aˆ . . . ø at centre = 2 % ø at circ.

& 1

Aˆ = 90º – x . . . ø sum of Δ

∴ 1

Mˆ = 2(90º – x)

= 180º – 2x �

10.3 AC is a diameter of ☼ACF . . . conv ø in semi-☼

& DACˆ = 90º . . . ø in semi-☼; diameter CD i.e. line AD ⊥ diameter AC

∴ AD is a tangent to ☼ACF � . . . conv tan ⊥ rad

10.4 In ΔEAB:

F is the midpoint of EA

& M is the midpoint of EB . . . line from centre of ☼

∴ FM = 1

2AB = 12 units . . . midpt theorem

& EB = DC = 2(6 + 12) . . . diameter = 36 units

∴ In right ød ΔEAB:

AE2 = EB2 – AB2 . . . Theorem of Pythagoras

= 362 – 242

= 720

∴ AE = 720 = 12 5 26,83 units �

OR: ÊAB = 90º . . . ø in semi-☼

In ΔEFM & ΔEAB

(1) Ê is common

(2) ÊFM = ÊAB

∴ ΔEFM | | | ΔEAB . . . øøø

∴ FM

AB =

EF

EA =

1

2

∴ FM = 12 units

∴ ME = 6 + 12 = 18 units . . . radii equal

∴ EF2 = ME2 – FM2 . . . Theorem of Pythagoras

= 182 – 122

= 180

∴ EF = 6 5

∴ AE = 12 5 ≈ 26,83 units

OR: In ΔEAB:

F is the midpoint of EA

& M is the midpoint of EB . . . line from centre of ☼

∴ FM = 1

2AB = 12 units . . . midpt theorem

∴ MC = 18

∴ MD = 18 . . . radii equal

ΔCFA ||| ΔAFD . . . øøø

∴ FA

FD =

CF

AF

∴ AF2 = CF . FD

= 6 .30

= 180

∴ AF = 6 5

∴ AE = 12 5

OR: ˆ

2A = x . . . ø in semi-☼

∴ ˆ2

A = Cˆ

∴ AD is a tangent to ☼ACF � . . . conv tan chord thm

C E

F

A

B D

M 1

1

2

2

3

3

x


2021

11.1 In the diagram, chords DE, EF and DF are

drawn in the circle with centre O.

KFC is a tangent to the circle at F.

Prove the theorem which states that

DFKˆ = Eˆ . (5)

D

E

K F

O

C

MEMOS 11.1

Method 1

Given: ?O with tangent at F and chord FE subtending ˆD at the circumference.

RTP: ˆDFK = Ê Construction: radii OF and OD

Proof: Let ˆDFK = x

ÔFK = 90º . . . radius ⊥ tangent

∴ ÔFD = 90º – x

∴ ÔDF = 90º – x . . . øs opposite equal radii

∴ ˆDOF = 2x . . . sum of øs in Δ

∴ Ê = x . . . ø at centre = 2 % ø at circumference

â ˆDFK = Ê �

Method 2

Given: ?O with tangent at F and chord FE subtending ˆD at the circumference.

RTP: ˆDFK = Ê Construction: diameter FM; join ME Proof: ˆMFK = 90º . . . tangent ⊥ diameter

& ˆMEF = 90º . . . ø in semi-?

Then . . .

Let ˆDFK = x

â ˆMFD = 90º – x

â ˆMED = 90º – x . . . øs in same segment

â ˆDEF = x

â ˆDFK = ˆDEF �

D

E

K F

O

Cx

D

E

K F

O

Cx

M

The angle between a tangent to a circle and a chord drawn from the point of

contact is equal to the angle subtended by the chord in the alternate segment.

Draw radii and use 'ø at centre' theorem.

We use 2 'previous' facts involving right øs

tangent ⊥ diameter . . . so, draw a diameter!

ø in semi-? = 90º . . . so, join RK!

These proofs are logical

& easy to follow.


D

E

F

O

K

C


2021

11.2 In the diagram, PK is a tangent to the

circle at K. Chord LS is produced to P.

N and M are points on KP and SP

respectively such that MN || SK.

Chord KS and LN intersect at T.

11.2.1 Prove, giving reasons, that:

(a) 4

Kˆ = NMLˆ (4)

(b) KLMN is a cyclic

quadrilateral. (1)


ΔLKN ||| ΔKSM. (5)

11.2.3 If LK = 12 units and

3KN = 4SM, determine the

length of KS. (4)

11.2.4 If it is further given that

NL = 16 units, LS = 13 units

and KN = 8 units, determine,

with reasons, the length of LT. (4)

[23]

1

1

1

1

1

2

2

2

2 2

3

3

3

4 K

N

T

L

S

M

P


2021

11.2 In the diagram, PK is a tangent to the

circle at K. Chord LS is produced to P.

N and M are points on KP and SP

respectively such that MN || SK.

Chord KS and LN intersect at T.

11.2.1 Prove, giving reasons, that:

(a) 4

Kˆ = NMLˆ (4)

(b) KLMN is a cyclic

quadrilateral. (1)


ΔLKN ||| ΔKSM. (5)

11.2.3 If LK = 12 units and

3KN = 4SM, determine the

length of KS. (4)

11.2.4 If it is further given that

NL = 16 units, LS = 13 units

and KN = 8 units, determine,

with reasons, the length of LT. (4)

[23]

11.2.1 (a) 4

Kˆ = 1

Sˆ . . . tan chord thm

= NMLˆ � . . . corresp øs

; MN || SK

(b) 4

Kˆ = NMLˆ

∴ KLMN is a cyclic quad. � . . .

11.2.2 LKNˆ = 1

Mˆ . . . ext ø of cyclic quad. KLMN

= 2

Sˆ . . . corresp øs

; MN || SK

∴ In Δs LKN & KSM

(1) ˆLKN = 2

Sˆ

(2) ˆ

3N = ˆ 3M . . . øs

in same segment

∴ ΔLKN | | | ΔKSM � . . . øøø

11.2.3 ∴ KS

LK =

SM

KN . . . | | |Δ

s

3KN = 4SM

∴ 3

4 =

SM

KN

∴ KS

12 =

3

4

∴ KS = 36

4

= 9 units �

11.2.4 4SM = 3KN = 3 % 8 = 24

∴ SM = 6 units

In ΔLNM: LT

LN =

LS

LM . . . prop thm; MN || ST

∴ LT

16 =

13

13 + 6

∴ LT = 13

19 % 16

= 208

19

≈ 10,95 units �

converse

ext ø of c.q.

MEMOS

1

1

1

1

1

2

2

2

2 2

3

3

3

4

K

N

T

L

S

M

P

ABOUT TAS

Gr 12 Maths 2 in 1 offers:

The questions are designed to:

� transition from basic concepts through to the more challenging concepts

� include critical prior learning (Gr 10 & 11) when this foundation is required for mastering the

entire FET curriculum

� engage learners eagerly as they participate and thrive on their maths journey

� accommodate all cognitive levels

a UNIQUE 'question & answer method'

of mastering maths

'a way of thinking'

To develop . . .

� conceptual understanding

� procedural fluency & adaptability

� reasoning techniques

� a variety of strategies for problem-solving

The questions and detailed

solutions have been provided in

SECTION 1: Separate topics

It is important that learners focus on and master one topic at a time BEFORE

attempting 'past papers' which could be bewildering and demoralising. In this

way they can develop confidence and a deep understanding.

&

SECTION 2: Exam Papers

When learners have worked through the topics and grown fluent, they can then

move on to the exam papers to experience working a variety of questions in

one session, and to perfect their skills.

The TOPIC GUIDES will enable learners to continue mastering one topic

at a time, even when working through the exam papers.

PLUS, . . .

CHALLENGING

QUESTIONS & MEMOS

These questions are Cognitive Level 3 & 4

questions, diagnosed as such following poor

performance of learners in recent examinations.

EXTENSION

SECTION

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