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SCHOOL SECTION74

PART - I : CIRCLE - TANGENT

THEOREMA tangent at any point of a circle is perpendicular to the radius, throughthe point of contact.

In adjoining figure,line PQ is a tangent at A andSeg AO is the radius throughthe point of contact A.

seg OA line PQ.

EXERCISE - 2.1 (TEXT BOOK PAGE NO.50)

2. In the adjoining figure,point A is the centre of the circle.AN = 10 cm. Line NM is tangent at M.Determine the radius of the circleif MN = 5 cm. (2 marks)

Sol. In AMN,m AMN = 90º [Radius is perpendicular to the tangent]

AN² = AM² + MN² [By Pythagoras theorem] 10² = AM² + 5² [Given] 100 = AM² + 25 AM² = 100 – 25 AM² = 75

AM = 75

AM = 25 3

AM = 5 3 cm.

Radius of the circle is 5 3 cm.


3. In the adjoining figure,Q is the centre of circle and PMand PN are tangent segments tothe circle. If MPN = 40º circle,find MQN. (2 marks)

Sol. In MQNP,m MPN = 40º [Given]m PMQ = 90º

[Radius is perpendicular to the tangent]m PNQ = 90º

m MQN = 140º [Remaining angle]

O

P QA

A

M N

P

N

Q

M

40º

Circle2.

MT EDUCARE LTD. GEOMETRY

SCHOOL SECTION 75


4. As shown the adjoining figure,two concentric circles are givenand line AB is tangent to the smallercircle at T. Show that T is themidpoint of seg AB. (2 marks)

Construction : Let O be the centre of theconcentric circles and draw seg OT.

Proof : Line AB is tangent to the smaller circle at T. [Given] seg OT line AB .......(i) [Radius is perpendicular to the tangent]

With respect to the bigger circle,seg OT chord AB [From (i)]

AT = BT [Perpendicular from the centre of acircle to a chord bisects the chord]

T is the mid-point of seg AB.


1. In the adjoining figure,point P is centre of the circleand line AB is the tangent to thecircle at T. The radius of the circleis 6 cm. Find PB ifTPB = 60º. (2 marks)

Sol. In PTB,m TPB = 60º [Given]m PTB = 90º [Radius is perpendicular to tangent]

m PBT = 30º [Remaining angle] PTB is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem

PT =1

2 PB [Side opposite to 30º]

6 =1

2 PB [Given]

PB = 6 × 2

PB = 12 cm


6. In the adjoining figure,line AB is tangent to both thecircles touching at A and B.OA = 29, BP = 18, OP = 61then find AB. (3 marks)

Construction : Draw seg PM seg OA, A - M - O

Sol. In PBAM,m PBA = 90º [Radius is perpendicular to the tangent]m BAM = 90ºm PMA = 90º [Construction]

m MPB = 90º [Remaining angle]

O

BTA

P

A T B

O

A B

P

18

61

M29

GEOMETRY MT EDUCARE LTD.

SCHOOL SECTION76

PBAM is a rectangle [By definition] PB = AM = 18 units [Opposite sides of a rectangle]

OA = OM + AM [ A - M - O] 29 = OM + 18 OM = 29 – 18 OM = 11 units

In PMO,m PMO = 90º [Construction]

OP2 = OM2 + PM2 [By Pythagoras theorem] (61)2= (11)2 + PM2

3721= 121 + PM2

PM2 = 3721 – 121 PM2 = 3600 PM = 60 units

But, AB = PM [Opposite sides of a rectangle]

AB = 60 units

PROBLEM SET - 2 (TEXT BOOK PAGE NO.192)

1. In the adjoining figure, in a circle with centre P,a chord AB is parallel to a tangent and intersectsthe radius drawn from a point of contact atthe midpoint of the radius. If AB = 12,find the radius of the circle. (3 marks)

Sol.Let the radius of the circle be 2x units

PA = PF = 2x units .......(i) [Radii of same circle]E is the midpoint of seg PF [Given]

PE = 1

PF2

PE = 1

2x2 [From (i)]

PE = x units ......(ii)

line AB || line CD [Given]

On transversal PF,

PEA PFC [Converse of corresponding angle test]

But, m PFC = 90º [Radius is perpendicular to the tangent]

m PEA = 90º ......(iii)

seg PE chord AB [From (iii)]

AE = 1

× AB2

[The perpendicular drawn from the centre

of the circle to the chord bisects the chord]

AE = 1

122 [Given]

AE = 6 units ......(iv)In PEA,m PEA = 90º [From (iii)]

A

P

E B

FC D••


SCHOOL SECTION 77

O

lA

P

A

B

O

PA2 = PE2 + AE2 [By Pythagoras theorem] (2x)2 = x2 + 62

4x2 – x2 = 36 3x2 = 36 x2 = 12 x = 4 × 3

x = 2 3 [Taking square roots]

PA = 2 × 2 3 = 4 3 units

Radius of the circle is 4 3 units.

THEOREMA line perpendicular to the radius at its outer end is a tangent to thecircle.

In adjoining figure,line l is perpendicular to radius OAat its outer end A,

line l is a tangent.


5. Tangents to the circle with centre O, at the points A and B intersect atP, a circle is drawn with centre P passing through A, Prove that thetangent at A to the circle with centre P passes through O. (3 marks)

Proof : Consider circle with centre O,Line AP is a tangent to the circle at point A.

[Given] m OAP = 90º ......(i)

[ Radius is perpendicular to the tangent]Consider circle with centre P,line OA seg AP [From (i)]

line OA is a tangent at point A [A line perpendicular to theradius at its outer end is tangent

to the circle] Tangent at A to the circle with centre P passes through O.

THEOREM

Statement : The lengths of the two tangent segments to a circle drawn from an external point are equal. (4 marks) Given : (i) A circle with centre O.

(ii) P is a point in the exterior of the circle.(iii) Points A and B are the points of contact of the two tangents from P to the circle.

To Prove : PA = PB Construction : Draw seg OA, seg OB and seg OP. Proof : In PAO and PBO,

m PAO = m PBO = 90º[Radius is perpendicular to the tangent]Hypotenuse OP Hypotenuse OP [Common side]

seg OA seg OB [Radii of same circle] PAO PBO [By hypotenuse - side theorem]

seg PA seg PB [c.s.c.t] PA = PB

A

O

B

P

C


SCHOOL SECTION78


7. Explain what is wrong with the information marked on the followingfigures (without taking actual measurements) (2 marks)

Sol. (i)

In ABC,m A = 60º [Given]m C = 30ºm B = 90º

Line BC is a tangent to the circle at point B[A Line perpendicular to the

radius at its outer end is a tangent to the circle]But according to given figure Line BC is not a tangent.

(ii)

Take a point D on ray ACsuch that A - C - Dm OCD + m OCA = 180º [Linear pair axiom]

110º + m OCA = 180º m OCA = 180 – 110 m OCA = 70º ......(i)

In OAC, m OCA = 70º [From (i)]m COA = 20º [Given]

m OAC = 90º [Remaining angle] Line AC is a tangent [A Line perpendicular to the

radius at its outer end is a tangent to the circle]But according to the given figure Line AC is not a tangent.

(iii)

In OPE,m OPE = 90º [Given]

OE2 = OP2 + EP2 [By Pythagoras theorem] 152 = 92 + EP2

225 = 81 + EP2

EP2 = 225 – 81 EP2 = 144 EP = 12 Units [Taking square roots] EP = EK = 12 Units [The lengths of two tangent segments

from an external point to a circle are equal]But according to the given figure EK = 13 units.

C

3

A

B

60º

30º

4

5

O

A

D

C

110º

20º

E

K

O

P

9

15

13


SCHOOL SECTION 79


8. In the adjoining figure,are four tangents to a circle at thepoints A, B, C and D. These four tangentsform a parallelogram PQRS.If PB = 5 and BQ = 3 then find PS. (4 marks)

Sol. PA = PB = 5BQ = CQ = 3Let,AS = SD = xCR = DR = yPQRS is a parallelogram [Given]

PQ = SR [ Opposite sides of a parallelogramare congruent]

PB + BQ = SD + DR [P - B - Q and S - D - R] 5 + 3 = x + y x + y = 8 ......(i)

PS = QR [ Opposite sides of a parallelogramare congruent]

PA +AS = QC + CR [ P - A - S and Q - C - R] 5 + x = 3 + y x – y = 3 – 5 x – y = – 2 ......(ii)

Adding (i) and (ii)x + y + x – y = 8 + (– 2)

2x = 8 – 2 2x = 6 x = 3

PS = PA + AS [ P - A - S] PS = 5 + x PS = 5 + 3

PS = 8 units

EXERCISE 2.1 (TEXT BOOK PAGE NO.52)

12. In the adjoining figure,AB and AC are tangents drawnfrom A, and BA CA. Prove thatBACO is a square. (2 marks)

Proof :In BACO,m OBA = 90º [Radius is perpendicular to the tangent]m OCA = 90ºm BAC = 90º [Given]

m BOC = 90º [Remaining angle] BACO is a rectangle [By definition]

seg OB seg OC [Radii of the same circle] BACO is a square [A rectangle in which adjacent sides

are congruent is a square]

O C

B A

BP Q

AC

SD

R

5 3

[The lengths of the two tangentsegments to a circle drawn from anexternal point are equal]


SCHOOL SECTION80


9. In the adjoining figure,point A is a common point of contactof two externally touching circlesand line l is a common tangentto both circles touching at B and C.Line m is another common tangentat A and it intersects BC at D.Prove that (i) BAC = 90º(ii) Point D is the midpoint of seg BC. (4 marks)

Proof : In BDA,DB = DA .......(i) [The lengths of the two tangent

segments from an external pointto a circle are equal]

DBA DAB [Isosceles triangle theorem]Let,m DBA = m DAB = xº ......(ii)In DAC,DA = DC ......(iii) [The lengths of the two tangent


DAC DCA [Isosceles triangle theorem]Let,m DAC = m DCA = yº......(iv)m BAC = m DAB + m DAC [Angle Addition Property]

m BAC = (x + y)º ........(v) [From (ii) and (iv)]In ABC,m ABC + m ACB + m BAC = 180º

[ Sum of the measures of theangles of a triangle is 180º]

x + y + x + y = 180 [From (ii), (iv), (v) and B - D - C] 2x + 2y = 180 2(x + y) = 180

x + y =180

2 x + y = 90 m BAC = 90º [From (v)]

From (i) and (iii) we get,DB = DC

D is the midpoint of seg BC.

EXERCISE - 2.1 (TEXT BOOK PAGE NO.52)10. In the adjoining figure,

ABC is isosceles triangle withperimeter 44 cm. The base BCis of length 12 cm. Sides AB andAC are congruent. A circle touchesthe three sides as shown. Find thelength of a tangent segment fromA to the circle. (4 marks)

Sol.Let, AP = AQ = x ......(i)

BP = BR = y ......(ii)CR = CQ = z ......(iii)

B

A

m

lCD

P

A

Q

RB C

[The lengths of the two tangentsegments to a circle drawn froman external point are equal]


SCHOOL SECTION 81

AB = AP + PB [ A - P - B] AB = x + y .....(iv) [From (i) and (ii)]

Similarly, BC = y + z BC = 12 .....(v)

AC = x + z .....(vi)Perimeter of ABC = 44 cm [Given]

AB + BC + AC = 44 x + y + y + z + x + z = 44 [From (iv), (v), and (vi)] 2x + 2y + 2z = 44 2(x + y + z) = 44 x + y + z = 22 x + 12 = 22 [From (v)] x = 22 – 12 x = 10 AP = AQ = 10 cm [From (i)]

Length of a tangent segment from A to the circle is 10 cm.

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195)

14. Centre of the circle is O, tangent CAat A and tangent DB at B intersecteach other at point P. Bisectors ofCAB and DBA intersect atpoint Q on the circle. APB = 60º,prove that PAB QAB. (4 marks)

Proof : In PAB,PA = PB [The lengths of two tangent

segments from an external point to a circle are equal] PAB PBA .....(i) [Isosceles triangle theorem]

m PAB + m PBA + m APB = 180º [Sum of the measures ofangles of a triangle is 180º]

m PAB + m PAB + 60º = 180º [From (i) and given] 2 m PAB = 180 – 60 2 m PAB = 120 m PAB = 60º m PAB = m PBA = 60º ......(ii) [From (i)]

m PAB + m BAC = 180º [Linear pair axiom] 60 + m BAC = 180º [From (ii)] m BAC = 120º .....(iii)

Similarly, m ABD = 120º .....(iv)

m QAB =1

2m BAC [ ray AQ bisects BAC]

=1

2 × 120º [From (iii)]

m QAB = 60º ......(v)Similarly,

m QBA = 60º .....(vi)In PAB and QAB,PAB QAB [From (ii) and (v)]seg AB seg AB [Common side]PBA QBA [From (ii) and (vi)]

PAB QAB [By ASA test of congruence]

P

C

D

O

B

Q

A

60º

••

××


SCHOOL SECTION82


2. In the adjoining figure,O is the centre and seg AB isa diameter. At the point C onthe circle, the tangent CD is drawn.Line BD is a tangent to the circle atthe point B. Show that seg OD || chord AC.

(4 marks)

Construction : Draw seg OC.Proof : In OAC,

seg OA seg OC [Radii of same circle] OAC OCA [Isosceles triangle of theorem]

Let,m OAC = m OCA = x ........(i)

In OCD and OBD,seg OC seg OB [Radii of same circle]seg CD seg BD [The lengths of two tangent


seg OD seg OD [Common side] OCD OBD [By SSS test of congruence] COD BOD [c.a.c.t]

Let,m COD = m BOD = y ........(ii)m BOC = m COD + m BOD [Angle addition property]

m BOC = y + y [From (ii)] m BOC = 2y ........(iii)

BOC is an exterior angle of AOCm BOC = m OCA + m OAC [Remote interior angles theorem]

2y = x + x [From (i) and (iii)] 2y = 2x y = x m BOD = m OAC [From (i) and (ii)] BOD m BAC [A - O - B] seg OD ||chord AC [Corresponding angles test]


3. Two circles with centers P andQ touch each other at point Texternally. seg BD is a diameter ofthe circle with centre Q. line BA isa common tangent touching theother circle at A. Prove that thepoints D, T and A are collinear. (5 marks)

Construction : Draw line MN passing throughpoint T such that A - M - B and M - T - Nand draw seg BT.

Proof : In MAT,AM = MT [The lengths of the two tangent

segments to a circle drawn froman external point are equal]

MAT MTA [Isosceles triangle theorem]

C

A B

D

O

A

P

QT

D

BM

N


SCHOOL SECTION 83

Let,m MAT = m MTA = x ......(i)Similarly,m MBT = m MTB = y ......(ii)m ATB = m MTA + m MTB [Angle addition property]

m ATB = x + y ......(iii) [From (i) and (ii)]

In ATB,m TAB + m ABT + m ATB = 180º [Sum of the measures of angles of

a triangle is 180º] x + y + x + y = 180 [From (i), (ii) and (iii)] 2x + 2y = 180 2 (x + y) = 180 x + y = 90 m ATB = 90º ......(iv) [From (iii)]

m DTB = 90º ........(v)[Angle subtended by a semicircle]

Adding (iv) and (v),m ATB + m DTB = 90 + 90

m ATB + m DTB = 180ºAlso ATB and DTB are adjacent angles.

ATB and DTB form a linear pair [Converse of linear pair axiom] ray TA and ray TD are opposite rays Points D, T and A are collinear.


5. In a right angled triangle ABC, ACB = 90º acircle is inscribed in the triangle with radius r.a, b, c are the lengths of the sides BC, AC andAB respectively. Prove that 2r = a + b – c.

(4 marks)

Proof : Let the centre of the inscribed circle be ‘O’Let AP = AQ = x ........(i)CP = CR = y .......(ii)BR = BQ= z ......(iii)a + b – c = BC + AC – AB

a + b – c = CR + RB + AP + PC – (AQ + QB)[B - R - C, A - P - C , A - Q - B]

a + b – c = y + z + x + y – (x + z) [From (i), (ii) and (iii)] a + b – c = y + z + x + y – x – z a + b – c = 2y a + b – c = 2y a + b – c = 2CP ........(iv) [From (ii)]

In PCROm OPC = m ORC = 90º [Radius is perpendicular to tangent]m PCR = 90º [Given]

m POR = 90º [Remaining angle] PCRO is a rectangle [By definition] CP = OR ........(v) [Opposite sides of a rectangle] a + b – c = 2 OR [From (iv) and (v)] a + b – c = 2r

A

O

Q

P

C R B



SCHOOL SECTION84

C

O D B

A


6. In the adjoining figure, points P, B and Qare points of contact of the respectivetangents. line QA is parallel to line PC.If QA = 7.2 cm, PC = 5 cm, find theradius of the circle. (5 marks)

Construction : Draw seg OA and seg OC.Sol. AQ = AB = 7.2 cm .....(i)

CP = CB = 5 cm ....(ii)In OQA and OBA,m OQA = mOBA = 90º [Radius is perpendicular to tangent]Hypotenuse OA Hypotenuse OA [Common side]

seg OQ seg OB [Radii of same circle] OQA OBA [By Hypotenuse side theorem] AOQ AOB [c.a.c.t.]

Let,m AOQ = m AOB = xº .....(iii)m QOB = m AOQ + m AOB [Angle addition property]m QOB = x + x [From (iii)]

m QOB = 2xº ......(iv)Similarly,m COP = m COB = yº ......(v)m POB = 2yº .....(vi)m AOC = m AOB + m COB [Angle addition property]m AOC = x + y ....(vii) [From (iii) and (v)]m QOB + m POB = 180º [Linear pair axiom]

2x + 2y = 180 2 (x + y) = 180 x + y = 90º m AOC = 90º .....(viii)[From (vii)]

In AOC,m AOC = 90º [From (viii)]seg OB Hypotenuse AC [Radius is perpendicular to tangent]

OB2 = AB × BC [By property of geometric mean] OB2 = 7.2 × 5 [From (i) and (ii)] OB2 = 36 OB = 6cm [Taking square roots]

Radius of the circle is 6 cm.


11. In the adjoining figure,BC and BA are tangents to circle.Prove that OD is perpendicularbisector of AC, where O is thecentre of the circle. (3 marks)

Proof :

OA = OC ........(i) [Radii of same circle]BC = BA .......(ii) [The lengths of the two tangent

segments to a circle drawn from an external point are equal]

CP

OB

Q A



SCHOOL SECTION 85

O AT

PO A

Points O and B are equidistant from the end points A and C of seg AC.[From (i) and (ii)]

Points O and B lie on the perpendicular bisector of seg AC.[By perpendicular bisector theorem]

seg OB is the perpendicular bisector of seg AC. seg OD is the perpendicular bisector of seg AC. [ O - D - B]

THEOREM

If two circles are touching circles then the common point lies on theline joining their centres.

Externally touching circles :In the adjoining figure,two circles with centres Oand A are touching externally at point P. O - P - A

Internally touching circles :In the adjoining figure,two circles with centres Oand A are touching internally at point P. O - A - P


1. If two circles touch externally then show that the distance betweentheir centers is equal to the sum of their radii. (2 marks)

Given : Two circles with centres O and Atouch each other externally at point T.

To Prove : OA = OT + ATProof : O - T - A [If two circles are touching

circles then the common pointlies on the line joining their centres]

OA = OT + AT [ O - T - A]


2. If two circles with radii 8 and 3 respectively touch internally thenshow that the distance between their centers is equal to the differenceof their radii, find that distance. (2 marks)

Sol. Let two circles with centres O andA touch each other internally at point T.

O - A - T [If two circles are touchingcircles then the commonpoint lies a the line joiningtheir centres]

OT = OA + AT [O - A - T] OA = OT – AT

OT = 8 units, AT = 3 units [Given] OA = 8 – 3 OA = 5 units

The distance between the centres is 5 units.

A PO

OA T


SCHOOL SECTION86


5. Three congruent circles withcentres A, B and C and withradius 5 cm each, touch eachother in points D, E, G as shown in(i) What is the perimeter of ABC ?(ii) What is the length of side DE of DEF ? (3 marks)

Sol. A - D - BB - E - CA - F - CAD = DB = BE = EC = AF = FC = 5cm .......(i)

[Radii of congruent circles and given]AB = AD + BD [ A - D - B]

AB = 5 + 5 [From (i)] AB = 10 cm ......(ii)

Similarly, BC = 10 cm ......(iii)AC= 10 cm .......(iv)

Perimeter of ABC = AB + BC + AC= 10 + 10 + 10 [From (ii), (iii) and (iv)]

Perimeter of ABC = 30 cm

(ii) In ABC,D and E are mid-points of sides AB and BC respectively. [From (i)]

DE =1

2AC [By mid-point theorem]

DE =1

2× 10 [From (iv)]

DE = 5 cm.


3. Two circles with centers A, B are touching externally and a circle with centreC touches both externally. Suppose AB = 3 cm, BC = 3 cm, CA = 4 cm.Find the radii of all circles. (4 marks)

Given : Circles with centres A, B and Ctouch each other pairwise externallyat points P, Q and R respectively.AB = 3 cm, BC = 3cm, CA = 4cm.

To Find : Radii of the circles with centre A, B and C.Sol. A - P - B

B - Q - CA - R - C

Let,AP = AR = xBP = BQ = y [Radii of the same circle]CQ = CR = zAP + BP = AB [ A - P - B]

x + y = 3 ........(i)BQ + CQ = BC [B - Q - C]

y + z = 3 .......(ii)AR + CR = AC [ A - R - C]

x + z = 4 ......(iii)

[If two circles are touching circlesthen the common point lies on theline joining their centres]

A B

C

D

EF

[If two circles are touching circlesthen the common point lies on theline joining their centres]

A B

C

P

R

Q


SCHOOL SECTION 87

Adding (i), (ii) and (iii),x + y + y + z + x + z = 3 + 3 + 4

2x + 2y + 2z = 10 2(x + y + z) = 10 x + y + z= 5 ......(iv)

Substituting (i) in (iv),3 + z = 5

z = 5 – 3 z = 2

Substituting (ii) in (iv),x + 3 = 5

x = 5 – 3 x = 2

Substituting (iii) in (iv),y + 4 = 5

y = 5 – 4 y = 1

Radii of circles with centres A, B and C are 2 cm, 1 cm and 2 cm respectively.


8. In the adjoining figure,A and B are centers of twocircles touching each other at M.Line AC and line BD are tangents.If AD = 6 cm and BC = 9 cm then findthe length of seg AC and seg BD. (3 marks)

Sol. A - M - B [If two circles are touching circles thenthe common point lies on the linejoining their centres]

AM = AD = 6 cm ........(i) [Radii of the same circle]BM = BC = 9 cm ........(ii)AB = AM + MB [ A - M - B]

AB = 6 + 9 [From (i) and (ii)] AB = 15 cm ........(iii)

In ABC,m ACB = 90º [Radius is perpendicular to the tangent]

AB² = AC² + BC² [By Pythagoras theorem] 15² = AC² + 9² [From (ii) and (iii)] 225 = AC² + 81 AC² = 225 – 81 AC² = 144 AC = 12 cm [Taking square roots]

In ADB,m ADB = 90º [Radius is perpendicular to the tangent]

AB2 = AD2 + BD2 [By Pythagoras theorem] 152 = 62 + BD2 [From (i) and (iii)] 225 = 36 + BD2

BD2 = 225 – 36 BD2 = 189

BD = 9 21 BD = 3 21 cm. [Taking square roots]

The lengths of seg AC and seg BD are 12 cm and 3 21cm respectively.

D

A B

C

M


SCHOOL SECTION88


4. In the adjoining figure,points P and Q are the centers ofthe circles. Radius QN = 3, PQ = 9.M is the point of contact of the circles.Line ND is tangent to the larger circle.Point C lies on the smaller circle.Determine NC, ND and CD. (5 marks)

Construction : Draw seg CM and seg DPSol. Q - M - P [If two circles are touching circles then

the common point lies on the line joining their centre]QN = 3 units

[Given]PQ = 9 unitsPN = PQ + QN [ P - Q - N]

PN = 9 + 3 PN = 12 units

QN = QM = 3 units [Radii of the same circle] MN = 6 units [ Diameter is twice the radius]

PM + QM = PQ [ P - M - Q] PM + 3 = 9 [Given] PM = 9 – 3 PM = 6 units

PM = PD = 6 units [Radii of the same circle]In NDP,m NDP = 90º ......(i) [Radius is perpendiculer to the tangent]

NP² = ND² + PD² [By Pythagoras theorem] 12² = ND² + 6² 144 = ND² + 36 ND² = 144 – 36 ND² = 108

ND = 108

ND = 36 3 ND = 6 3 units

Now, m NCM = 90º .....(ii) [Angle inscribed in a semicircleis a right angle]

NCM NDP [From (i) and (ii)] seg CM || seg DP ....(iii) [By corresponding angles test]

In NDP,seg CM || seg DP, [From (iii)]

NC

CD=

NM

MP[By B.P.T.]

NC

CD=

6

6

NC

CD= 1

NC = CD C is the midpoint of seg ND

NC = CD = 1

2ND

NC = CD = 1

6 32

NC = CD = 3 3 units.

CD

Q PN M


SCHOOL SECTION 89


4. In the adjoining figure, two circles toucheach other internally in a point A. The radiusof the smaller circle with centre M is 5.The smaller circle passes through the centre Nof the larger circle. The tangent to the smallercircle drawn through C intersects the larger circlein point D. Find CD. (4 marks)

Construction : Draw seg AD.Sol. NM = MA = MP = 5 units [Radii of same circle]

NA = 2 NM [Diameter is twice the radius] NA = 2 × 5 NA = 10 units

CN = NA = 10 units [Radii of same circle]CA = 2 CN [Diameter is twice the radius]CA = 2 × 10CA = 20 unitsCM = CN+ NM [C - N - M]

CM = 10 + 5 CM = 15 units

m CPM = 90º [Radius is perpendicular to tangent]m CDA = 90º [Angle subtended by a semicircle]

CPM CDA seg PM || seg DA ......(i) [By corresponding angles test]

In CPMCPM = 90º [Radius is perpendicular to the tangent]CM2 = CP2 + PM2 [By Pythagoras theorem]152 = CP2 + 52

225 = CP2 + 25 CP2 = 225 – 25 CP2 = 200

CP = 100 × 2 [Taking square roots] CP = 10 2 units

In CDA,seg PM || seg DA [From (i)]

CP

PD=

CM

MA[By B.P.T.]

10 2

PD=

15

5

PD =10 2 × 5

15

PD =10 2

3units

CD = CP + PD [C - P - D]

CD =10 2 10 2

1 3

CD =30 2 10 2

3

CD =40 2

3 units

A

DP

MNC


SCHOOL SECTION90

EXERCISE - 2.2 (TEXT BOOK PAGE NO. 56)

7. Explain what is wrong in theinformation provided in theadjacent figure (do not takeactual measurements).OP = 4, PB = 4, BQ = 5. (2 marks)

Sol. Let the two circles touch at point M.O - M - Q [If two circles are touching circles

then the common point lies on the joining their centres]OM = OP = 4 units

[Radii of same circle]MQ = BQ = 5 units OQ = OM + MQ [O - M - Q] OQ = 4 + 5 OQ = 9 units

In OBQm OBQ = 90º [Radius is perpendicular to tangent]

OQ2 = OB2 + BQ2 [By Pythagoras theorem] OQ2 = 82 + 52

OQ2 = 64 + 25 OQ2 = 89

OQ = 89 units [Taking square roots]

But one segment cannot have two different lengths.


6. Two circles which are not congruent touch externally. The sum of theirareas is 130cm2 and the distance between their centers is 14 cm. Findradii of circles. (3 marks)

Sol. Let the radius of first circle be r1 and that of second circle be r2

Circles are touching externally r1 + r2 = 14 cm r2 = 14 – r1 ........(i)

According to given informationA (I Circle) + A (II Circle) = 130 cm2

r12 + r2

2 = 130 (r1

2 + r22) = 130

r12 + r2

2 = 130 r1

2 + (14 – r1)2 = 130 [From (i)]

r12 + 196 – 28r

1 + r1

2 = 130 2r1

2 – 28r1 + 196 – 130 = 0 2r1

2 – 28r1 + 66 = 0 2 (r1

2 – 14r1 + 33) = 0 r1

2 – 14r1 + 33 = 0 r1

2 – 11r1 – 3r1 + 33 = 0 r1 (r1 – 11) – 3 (r1 – 11) = 0

(r1 – 3) (r1 – 11) = 0r1 – 3 = 0 or r1 – 11 = 0

r1 = 3 units or r1 = 11 unitsIf r1 = 3 If r1 = 11

then,r2 = 14 – 3 then, r2 = 14 – 11r2 = 11 units r2 = 3 units

The radii of the circles is 11 units and 3 units.

B

Q

P

O

4

4


SCHOOL SECTION 91

PART - II : CIRCLE - ARC

• An arc is a part of a circle.

• If a secant passes through the centre,then each arc so formed is called a semicircle.In the adjoining figure, arc AXB and arc AYBare semicircles.

• An arc smaller than the semicircleis called a minor arc.

• An arc greater than the semicircle is called a major arc.In the adjoining figure, arc AXB is a minor arcand AYB is the corresponding major arc.

• An angle in the plane of the circle with its vertexat the centre is called a central angle.

In the adjoining figure,AOB is the central angle correspondingto minor arc AXB.

• Measure of a minor arc is equal to the measureof its corresponding central angle.In the adjoining figure, m (arc AXB) = m AOB

• Measure of major arc = 3600 – measure of corresponding minor arc.

In the adjoining figure,m(arc AYB) = 3600 – m(arc AXB)

• Measure of a circle is 3600

• Measure of a semicircle is 1800

` Arc addition property :In the adjoining figure,arc APB and arc BQC have incommon only the end point B,\ m(arc APB) + m(arc BQC) = m(arc ABC)

` Congruent arcs :Two arcs of the same circle or of congruent circles, having equal measuresare congruent.

` Inscribed angle :An angle is said to be an inscribed angle, if(i) the vertex is on the circle and(ii) both the arms are secants.

In the adjoining figure,ABC is an inscribed angle,because vertex B lies on thecircle and both the arms BAand BC arc secants.In other words, ABC is inscribed in arc ABC.

A•

BO

Y

X•

•

A X

O•

Y

B• •

•

•

•

BA

O

(

X•

A

Y

•

X

B

•

••

•

O

•Q

A

PB

C

•

• •

•

•

O

•

CA

B


SCHOOL SECTION92

(a) (c)(b)

B

A

C

D

E

DB

A

C

A

C

B

(d) (f)

A

B

C

A

B

C

(e)

A

B

C

INSCRIBED ANGLE THEOREM

B

AC

X•

` Intercepted arc :Given an arc of the circle and an angle, if each side of the angle contains anend point of the arc and all other points of the arc except the end points liein the interior of the angle, then the arc is said to be intercepted by the angle.

(i) In figures (a), (b) and (c), ABC has its vertex B outside the circle andintercepts two arcs.

(ii) In figures (d) and (e), ABC has its vertex on the circle and interceptsonly one arc.

(iii) In figure (f), ABC has its vertex B inside the circle and intercepts onlyone arc.

The measure of an inscribed angle is half of the measure of itsintercepted arc.

In the adjoining figure,ABC is an inscribed angle andarc AXC is the intercepted arc.

m ABC = 1

2 m (arc AXC)

Corollary - 1 :An angle inscribed in a semicircle is a right angle.In the adjoining figure,ABC is inscribed inthe semicircle ABC.

m ABC = 90º

•

B

AO

C


SCHOOL SECTION 93

B

A

D

C•X

Corollary - 2 :Angles inscribed in the same arc are congruent.

In the adjoining figure,ABC and ADC, both areinscribed in the same arc ABC.

ABC ADC.

NOTE Inscribed angles, ABC and ADC both intercept the same arc AC.

ABC ADC.


1. In the adjoining figure,if m (arc APC) = 60ºand m BAC = 80ºFind (a) ABC (b) m (arc BQC). (2 marks)

Sol.

(a) m ABC =1

2 m(arc APC) [Inscribed angle theorem]

m ABC =1

2 × 60

m ABC = 30º

(b) m BAC =1

2 m(arc BQC) [Inscribed angle theorem]

80 =1

2 m(arc BQC)

m(arc BQC) = 80 × 2

m(arc BQC) = 160º


3. Chords AB and CD of a circleintersect in point Q in the interiorof a circle as shown in figure,if m (arc AD) = 25º circle andm (arc BC) = 31º, then find BQC. (2 marks)

Construction : Draw seg AC.Sol. m (arc AD) = 25º [Given]

m ACD =1

2 m(arc AD) [Inscribed angle theorem]

m ACD =1

2 × 25º

m ACD = 12.5º m ACQ = 12.5º [ D - Q - C]

m (arc BC) = 31º [Given]

B

A

C80º

•Q

•P

D B

QAC


SCHOOL SECTION94

m BAC =1

2 m(arc BC) [Inscribed angle theorem]

m BAC =1

2 × 31º

m BAC = 15.5º m QAC = 15.5º [ A - Q - B]

BQC is an exterior angle of AQC, m BQC = m QAC + m ACQ [Remote interior angle theorem] m BQC = 15.5º + 12.5º

m BQC = 28º


4. Secants containing chords RSand PQ of a circle intersects eachother in point A in the exterior ofa circle, as shown in figure Ifm (arc PCR) = 26º andm (arc QDS) = 48ºthen find(i) AQR (ii) SPQ (iii) RAQ (3 marks)

Sol. (i) m (arc PR) = 26º [Given]

m AQR =1

2 m (arc PCR) [Inscribed angle theorem]

m AQR =1

2 × 26º [Given]

m AQR = 13º

(ii) m (arc QS) = 48º [Given]

m SPQ =1

2 m (arc QDS) [Inscribed angle theorem]

m SPQ =1

2 × 48º [Given]

m SPQ = 24º

(iii) m SRQ =1

2m (arc QDS) [Inscribed angle theorem]

m SRQ =1

2× 48º [Given]

m SRQ = 24º

SRQ is an exterior angle of ARQ,

m SRQ = m RAQ + m AQR [Remote interior angle theorem]

24º = m RAQ + 13º

m RAQ = 24º – 13º

m RAQ = 11º

D•

S

Q

C

R

P

A


SCHOOL SECTION 95


5. In the adjoining figure,in two chords AB and CDof the same circle are parallelto each other. P is the centreof the circle.Prove : m CPA = m DPB. (3 marks)

Construction : Draw seg BC.Proof : m CPA = m (arc CA) .......(i) [Definition of measure of minor arc]

m DPB = m (arc DB) ......(ii)

m ABC = 1

2 m (arc CA) ......(iii)

[Inscribed angle theorem]m BCD =

1

2 m (arc DB) ......(iv)

chord CD || chord AB [Given] On transversal BC,

ABC BCD ......(v) [Converse of alternate angles test]

1

2m (arc CA) =

1

2 m(arc DB) [From (iii), (iv) and (v)]

m (arc CA) = m (arc DB) ......(vi) m CPA = m DPB [From (i), (ii) and (vi)]


6. In the adjoining figure,m (arc XAZ) = m (arx YBW).Prove that XY || ZW

(2 marks)

Construction : Draw seg XW

Proof : m XWZ = 1

2 m (arc XAZ) .......(i)

[Inscribed angle theorem]m WXY =

1

2 m (arc YBW) .......(ii)

But, m (arc XAZ) = m (arc YBW) .......(iii) [Given] m XWZ = m WXY [From (i), (ii) and (iii)] line XY || line ZW [Alternate angles test]


7. Given circle with centre Oand BC || ED, m (arc BC) = 94º,m (arc ED) = 86º, ADE = 8ºFind (i) m (arc AE), (ii) m (arc DC),(iii) m (arc EB).Also find DAB, ECB, CBE. (5 marks)

Construction : Draw seg BD

Proof : m BDE = 1

2 m (arc BAE) .....(i)

[Inscribed angle theorem]

m CBD = 1

2 m (arc CD) .....(ii)

line BC || line ED [Given]

C DA B

P

X Y

A B

Z W

B C

A

E D


SCHOOL SECTION96

On transversal BDCBD BDE ......(iii) [Converse of alternate angles test]

1

2 m (arc BAE) =

1

2 m (arc CD) [From (i), (ii) and (iii)]

m (arc BAE) = m (arc CD) .......(iv)

m ADE =1

2 m (arc AE) [Inscribed angle theorem]

8º =1

2 m (arc AE)

m (arc AE) = 16ºm (arc AD) = m (arc AE) + m (arc ED) [Arc Addition property]

86º = m (arc ED) + 16º m (arc ED) = 86 – 16 m (arc ED) = 70º .......(iii)

m (arc BC) + m (arc CD) + m (arc BAE) + m (arc ED) = 360º[Measure of a circle is 360º]

94º + m (arc BAE) + m (arc BAE) + 70º = 360º[From (ii), (iii) and given]

2 m (arc BAE) + 164 = 360 2 m (arc BAE) = 360 – 164 2 m (arc BAE) = 196

m (arc BAE) = 98º ........ (v)

m (arc DC) = 98º[From (iv) and (v)]

m (arc BCD) = m (arc BC) + m (arc CD) [Arc Addition property]= 94 + 98

m (arc BCD) = 192º

m DAB =1

2 m (arc BCD) [Inscribed angle theorem]

m DAB =1

2 × 192º

m DAB = 96º

m ECB =1

2 m (arc BAE) [Inscribed angle theorem]

=1

2 × 98º

m ECB = 49º

m (arc CDE) = m (arc CD) + m (arc DE) [Arc addition property]= 98 + 70

m (arc CDE) = 168º

m CBE =1

2 m (arc CDE) [Inscribed angle theorem]

m CBE =1

2 × 168º

m CBE = 84º


SCHOOL SECTION 97


2. Find the radius of the circle passing through the vertices of a right angledtriangle when lengths of perpendicular sides are 6 and 8. (3 marks)

Given : (i) In PQR, m PQR = 90º(ii) PQ = 6 units, QR = 8 units.(iii) Points P, Q and R lie on the circle.

To Find : radius of the circle.Sol. In PQR,

m PQR = 90º [Given] PR² = PQ² + QR² [By Pythagoras theorem] PR² = 6² + 8² PR² = 36 + 64 PR² = 100 PR = 10 units [Taking square roots]

Let Y be a point on the circle as shown in the figurem PQR = 90º [Given]

m PQR = 1

2 m(arc PYR) [Inscribed angle theorem]

90º =1

2m(arc PYR)

m (arc PYR) = 180º arc PYR is a semicircle seg PR is the diameter. Diameter = 10 units. Radius = 5 units [ Radius is half of the diameter]

Radius of the circle is 5 units.


1. Let two circles intersect each at points A and D. Let the diameter ABintersect the circle with centre P at point N and diameter AC intersectsthe circle in point M with centre Q. Then prove that AC. AM = AB. AN.

(3 marks)

Construction : Draw seg BM and seg CN

Proof : seg AB is the diameter

m AMB = 90º [Angle subtended by a semicircle]

seg AC is the diameter

m ANC = 90º [Angle subtended by a semicircle]

m AMB = mANC .......(i)

In AMB and ANC

AMB ANC [From (i)]

BAM CAN [Common angle]

AMB ~ ANC [By AA test of similarity]

AM

AN =

AB

AC[c.s.s.t]

AC.AM = AB.AN

8Y

R

Q

P

6

•

A

Q PN

M

B CD


SCHOOL SECTION98

C

AD

B


11. ABC is inscribed in a circle with centre O, seg AXis a diameter of the circle with radius r.seg AD seg BC. Prove that

(i) ABX ~ ADC, (ii) A (ABC) = abc4r

.(a is side opposite to A, ...) (4 marks)

Proof : seg AX is a diameter [Given] arc ACX is a semi circle. m ABX = 90º [Angle subtended by a semicircle]

In ABX and ADCABX ADC [Each is 90º]AXB ACD [Angles inscribed in the same arc

and C - D - B] ABX ~ ADC [By AA test of similarity]

BC = a, AB = c, AC = b [Given]

AB

AD=

AX

AC[c.s.s.t.]

c

AD=

2r

b

AD =bc

2r.......(i)

A (ABC) =1

2 × base × height

=1

2 × BC × AD

=1

2 × a ×

bc

2r

A (ABC) =abc4r

Cyclic Quadrilateral :A quadrilateral whose all the four vertices lie on acircle is called cyclic quadrilateral.In the adjoining figure,ABCD is cyclic,as all the four vertices A, B, C and D.lie on a circle.

THEOREM Statement : The opposite angles of a cyclic quadrilateral are supplementary. Given : ABCD is a cyclic To Prove : m ABC + m ADC = 180º

m BAD + m BCD = 180º

Proof: m ABC = 1

2 m (arc ADC) .....(i)

m ADC = 1

2 m (arc ABC) .....(ii)

Adding (i) and (ii), we getC

DB

A

[Inscribed angletheorem]

A

B D C

O

X


SCHOOL SECTION 99

m ABC + m ADC = 1

2 m (arc ADC) +

1

2m (arc ABC)

m ABC + m ADC = 1

2[m (arc ADC) + m (arc ABC)]

m ABC + m ADC = 1

2 × 360º [ Measure of a circle is 360º]

m ABC + m ADC = 180º ..........(iii)In ABCD,m BAD + m BCD + m ABC + m ADC = 360º [ Sum of measure of

angles of a quadrilateral is 360º] m BAD + m BCD + 180º = 360º [From (iii)] m BAD + m BCD = 180º

THEOREM : If opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.


7. If two consecutive angles of cyclic quadrilateral arecongruent, then prove that one pair of opposite sidesis congruent and other is parallel. More precisely.Given : ABCD is cyclic quadrilateral in whichABC BCD. To prove side DC side AB, AD || BC.

(3 marks)Given : ABCD is cyclic and ABC BCDTo Prove : (i) seg AD || seg BC (ii) side DC side AB.Construction : Draw seg AC and seg BDProof : ABCD is cyclic [Given]

m ABC + m ADC =180º .......(i) [Opposite angles of a cyclicquadrilateral are supplementary]

But, ABC BCD .......(ii) [Given] m BCD + mADC = 180º [From (i) and (ii)] seg AD || seg BC [Interior angles test]

In ABC and DCBseg BC seg BC [Common side]ABC DCB [Given]BAC CDB [Angles inscribed in same arc]

ABC DCB [By SAA test] side AB side DC [c.s.c.t.]


19. Prove that in a cyclic trapezium angles at the base are congruent.(3 marks)

Given : ABCD is a cyclic trapezium.seg AB || seg BC

To prove : ABC DCBProof : ABCD is cyclic [Given]

m ABC + m ADC = 180º .....(i) [Opposite angles of a cyclicquadrilateral are supplementary]

seg AD || seg BC [Given] On transversal DC,

m DCB + m ADC = 180º ......(ii) [Converse of interior angle test]

B

A D

C

A D

B C


SCHOOL SECTION100

AD

B C

B

DA

C E•

m ABC + m ADC = m DCB + m ADC [From (i) and (ii)] m ABC = m DCB ABC DCB


18. In a cyclic quadrilateral ABCD,the bisectors of opposite angles Aand C meet the circle at P and Qrespectively. Prove that PQ is adiameter of the circle. (4 marks)

Proof :DAP BAP [ ray AP bisects DAB]Let m DAP = m BAP = xº ....(i)DCQ BCQ [ ray CQ bisects DCB]Let, m DCQ = m BCQ = yº .....(ii)ABCD is cyclic [Given]

m DAB + m DCB = 180º [Opposite angles of a cyclicquadrilateral are supplementary]

m DAP + m BAP + DCQ + m BCQ = 180º[Angle addition property]

x + x + y + y = 180º [From (i) and (ii)] 2x + 2y = 180º .....(iii)

m DAP =1

2 m (arc DP) [Inscribed angle theorem]

x =1

2m (arc DP) [From (i)]

m (arc DP) = 2xº .....(iv)

m DCQ =1

2 m (arc DQ) [Inscribed angle theorem]

y =1

2 m (arc DQ) [From (ii)]

m (arc DQ) = 2yº ......(v)m (arc DP) + m (arc DQ) = 2x + 2y [Adding (iv) and (v)]

m (arc PDQ) = 180º [Arc addition property and from (iii)] Arc PDQ is a semicircle seg PQ is a diameter of the circle.

Converse of cyclic quadrilateral theorem :If the opposite angles of a quadrilateral are supplementarythen it is a cyclic quadrilateral.In ABCD if m A + m C = 180ºorm B + m D = 180ºthen, ABCD is cyclic quadrilateral.

Corollary :An exterior angle of cyclic quadrilateral is congruent to the angle oppositeto adjacent interior angle.In the adjoining figure,ABCD is cyclicDCE is an exterior angle.DCE BAD

D P

Q

A

C

B

••

××


SCHOOL SECTION 101

NOTE Angle opposite to adjacent interior angle is also called interior opposite angle.


13. Prove that the quadrilateral formed by theangle bisectors of a cyclic quadrilateralis also cyclic. (5 marks)

Given : (i) ABCD is cyclic.(ii) Ray AP, ray BQ, ray CR and

ray DS are the bisectors ofA, B, C and D respectively.

To Prove : PQRS is cyclic.Proof : DAP BAP [ray AP bisects BAD]

Let, m DAP = m BAP = aº ........(i)Similarly,m ABP = m CBP = bº ........(ii)m BCR = m DCR = cº .......(iii)m ADR = m CDR = dº ........(iv)In BQC,m BQC + m QBC + m QCB = 180º [Sum of the measures of

angles of a triangle is 180º] m BQC + b + c = 180 [From (ii) and (iii)] m BQC = (180 – b – c)º m PQR = (180 – b – c)º .......(v)

[B - P - Q and C - R - Q]Similarly, we can prove

m PSR = (180 - a - d)º ......(vi)Adding (v) and (vi),

m PQR + m PSR = 180 – b – c + 180 – a – d m PQR + m PSR = 360 – a – b – c – d m PQR + m PSR = 360 – (a + b + c + d) .......(vii)

In ABCD,m BAD + m ABC + m BCD + m ADC = 360º

[ Sum of the measures of angles of a quadrilateral is 360º]m BAP + m DAP + m ABP + m PBC + = 360m BCQ + m DCQ + m CDR + m ADR

[Angle addition property] a + a + b + b + c + c + d + d = 360

[From (i), (ii), (iii) and (iv)] 2a + 2b + 2c + 2d = 360 2 (a + b + c + d) = 360 a + b + c + d = 180 ......(viii) m PQR + m PSR = 360 – 180

[From (vii) and (viii)] m PQR + m PSR = 180º PQRS is cyclic. [If opposite angles of a quadrilateral

are supplementary, then quadrilateral is cyclic]

•

××

B

C

D

A

SP

QR

•


SCHOOL SECTION102

A

P Q

B S R C


15. If all the sides of a quadrilateral are produced in the same order (clockwiseor anticlockwise). Show that quadrilateral formed by the bisectors ofexterior angles is cyclic. (5 marks)

Given : (i) ABCD is cyclic(ii) Lines PQ, QR, RS and PS arethe bisectors of the LAB, MBC, NCD,KDA respectively.

To prove : PQRS is cyclicProof : Let m LAQ = m BAQ = aº .......(i)

m MBR = m CBR = bº ......(ii)m NCS = m DCS = cº .....(iii)m KDP = m ADP = dº ......(iv)

[Lines PQ, QR RS and PS arethe bisectors]

m SPQ = 180 – a – dSimilarly,m QRS = 180 – b – c ......(vi)m PQR = 180 – a – b .....(vii)m PSR = 180 – c – d ....(viii)Adding (v) and (vi),

m SPQ + m QRS = 180 – a – d + 180 – b – cm SPQ + m QRS = 360 – (a + b + c + d) ......(ix)

In PQRS,m SPQ + m PQR + m QRS + m PSR = 360º

[Sum of the measures of angles of quadrilateral is 360º] 180 – a – d + 180 – a – b + 180 – b – c + 180 – c – d = 360 720 – 2a – 2b – 2c – 2d = 360 720 – 360 = 2a + 2b + 2c + 2d 360 = 2a + 2b + 2c + 2d a + b + c + d = 180º .....(x)

[Dividing throughout by 2] m SPQ + m QRS = 360º – 180º

[From (ix) and (x)] m SPQ + m QRS = 180º PQRS is cyclic [If the opposite angles of quadrilateral

are supplementary then it is cyclic]


21. In ABC, midpoints of sides AB,AC and BC are P, Q and R respectively.AS BC. Prove that PQRS is acyclic quadrilateral. (5 marks)

Proof : In ABC,P and Q are midpoint of seg AB and seg AC

seg PQ || seg BC [By midpoint theorem] seg PQ || seg BR [B - R - C]

Q

A B

R

C

S

KD

M

L

N

P

××

••

(

(


SCHOOL SECTION 103

Similarly, seg QR || seg PB PBRQ is a parallelogram [By definition]

In ASB,m ASB = 90º [Given]seg SP is median to hypotenuse AB [Given]

SP =1

2AB .....(i) [In a right angled triangle the median

drawn to the hypotenuse is half of it]

But, PB =1

2AB .....(ii) [ P is the midpoint of side AB]

In PBS,SP = PB [From (i) and (ii)]

m PBS = m PSB [Isosceles triangle theorem]PBRQ is a parallelogram

m PBR = m PQR [Opposite angles of a parallelogramare congruent]

m PBS = m PQR .....(iv) [B - S - R] m PSB = m PQR ......(v) [From (iii) and (iv)]

But,m PSB + m PSR = 180º [Linear pair axiom]m PQR + m PSR = 180º [From (v)]

PQRS is cyclic [If opposite angles of a quadrilateral aresupplementary then it is a cyclic quadrilateral]


16. In the adjoining figure,in the isosceles triangle PQR,the vertical P = 50º. The circlepassing through Q and R cuts PQ .in S and PR in T. ST is joined.Find PST. (3 marks)

Sol.In PQRseg PQ seg PR [Given]

PQR PRQ ......(i) [Isosceles triangle theorem]m PQR + m PRQ + m QPR = 180º [Sum of the measures of angles

of a triangle is 180º] m PRQ + m PRQ + 50 = 180º[From (i) and Given]

2m PRQ = 180º – 50º 2m PRQ = 130º m PRQ = 65º ......(ii)

SQRT is cyclic PST TRQ [An exterior angle of cyclic

quadrilateral is congruent to theangle opposite to adjacent interiorangle]

PST PRQ [P - T - R]

PST = 65º [From (ii)]

P

ST

50º

Q R


SCHOOL SECTION104

C AD

M B N


17. ABCD is a parallelogram.A circle passing through D, A,B cuts BC in P.Prove that DC = DP. (3 marks)

Proof :

ABPD is cyclic [By definition] DPC DAB .....(i) [An exterior angle of cyclic quadrilateral

is congruent to the angle opposite toadjacent interior angle]

ABCD is parallelogram DCB DAB .......(ii) [Opposite angles of a parallelogram

are congruent] DPC DCB ......(iii) [From (i) and (ii)]

In DPC,DPC DCP [From (ii) and C - P - B]

seg DP seg DC [Converse of isosceles triangle theorem] DP = DC


8. Two circles intersect each other in points X and Y. Secants through Xand Y intersect one of the circles in A and D and the other in B and Crespectively. A and B are on opposite sides of line DC. Show that lineAD is parallel to line BC. (3 marks)

Proof : ADYX is cyclic [By definition] XYC DAX ......(i) [The exterior angle of a cyclic

quadrilateral is equal to its interioropposite angle]

But, XYC XBC .....(ii) [Angles inscribed in the same arc] DAX XBC [From (i) and (ii)] DAB ABC [A - X - B] line AD || line BC [By Alternate angles test]


20. Two circles intersect each otherin points A and B. Secants throughA and B intersects circles in C, D andM, N. Prove that CM || DN. (3 marks)

Construction : Draw seg AB.Proof : ABMC is cyclic [By definition]

m MCA + m MBA = 180º ......(i) [Opposite angles of a cyclicquadrilateral are supplementary]

ABND is cyclic [By definition]

C

B

PD

A

A

DY

XC

B


SCHOOL SECTION 105

MBA ADN .....(ii) [The exterior angle of a cyclicquadrilateral is equal to its interioropposite angle]

m MCA + m ADN = 180º [From (i) and (ii)] m MCD + m CDN = 180º [C - A - D] seg CM || seg DN [By Interior angles test]


12. ABCD is cyclic quadrilateral, lines AB and DCintersect in the point F and lines AD and BCintersect in the point E. Show that thecircumcircles of BCF and CDE intersectin a point G on the line EF. (4 marks)

Construction : Draw seg CG.Proof : Let the circumcircle of BCF and CDE intersect at points C and G.

BCGF is cyclic. [By definition] m ABC = m CGF .....(i) [Exterior angle of a cyclic quadrilateral

is equal to its interior opposite angle]DCGE is cyclic [By definition]

m ADC = m CGE ......(ii) [Exterior angle of a cyclic quadrilateralis equal to its interior opposite angle]

Adding (i) and (ii),m ABC+ m ADC = m CGF + m CGE .......(iii)ABCD is cyclic

m ABC + m ADC = 180º ....(iv) [Opposite angle of a cyclic quadrilateralare supplementary]

m CGF + m CGE = 180º [From (iii) and (iv)]Also, CGF and CGE are adjacent angle

CGF and CGE form a linear pair [Converse of linear pair axiom] ray GF and ray GE form opposite rays Points F, G, E are collinear. G lies on line EF The circumcircle of BCF and CDE intersect in a point on line EF.


10. In the adjoining figure,two circles intersects each otherin points A and B. Secants through thepoint A intersect the circles in points P, Qand R, S. Line PR and line SQ intersect in point T.Show that(i) PTQ and PBQ are supplementary,(ii) BSTR is a cyclic quadrilateral. (5 marks)

Construction : Draw seg ABProof :(i) PBAR is cyclic [By definition]

TRA ABP [The exterior angle of a cyclic quadrilateralis equal to its interior opposite angle]

TRS ABP ......(i) [R - A - S]

FG

C

E

B D

A

T

SP

R Q

B

A


SCHOOL SECTION106

B

A

C

X

ABQ ASQ [Angles inscribed in same arc] ABQ RST .....(ii) [R - A - S and S - Q - T]

In TRS,m TRS + m RST + m RTS = 180º [Sum of measures of angles of

a triangle is 180º] m ABP + m ABQ + m RTS = 180º [From (i) and (ii)] m PBQ + m RTS = 180º [Angle addition property] m PBQ + m PTQ = 180º [Q P - R - T and T - Q - S]

(ii)

ABR APR [Angles inscribed in the same arc]ABR QPT ......(i) [P - R - T and P - A - Q]ABSQ is cyclic [By definition]

TQA ABS [The exterior angle of a cyclic quadrilateralis equal to its interior opposite angle]

TQP ABS ......(ii) [P - A - Q]In TPQ,m QPT + m TQP + m PTQ = 180º [Sum of measures of angles of

a triangle is 180º] m ABR + m ABS + m PTQ = 180º [From (i) and (ii)] m RBS + m PTQ = 180º [Angle addition property] m RBS + m RTS = 180º [P - R - R and T - Q - S] BSTR is cyclic [If opposite angles of a quadrilateral

are supplementary then thequadrilateral is cyclic]

TANGENT SECANT THEOREMIf an angle with its vertex on the circle whose one side touches thecircle and the other intersects the circle in two points, then themeasure of the angle is half the measure of its intercepted arc.In the adjoining figure,ABC has its vertex B on thecircle, line BC is tangent to circleat B and ray BA is a secant.

m ABC = 1

2m(arc AXB)

NOTE If an angle,(i) has its vertex on the circle.(ii) one arm is a tangent, and(iii) the other arm is a secant, then, we will term such an angle as a

Tangent secant angle.

T

SP

R Q

B

A


SCHOOL SECTION 107


1. In the adjoining figure,seg AB and seg AD are chordsof the circle. C be a point ontangent to the circle at point A.If m (arc APB) = 80º and BAD = 30º,then find (i) BAC (ii) m (arc BQD) (2 marks)

Sol. m BAC =1

2 m(arc APB) [Tangent secant theorem]

m BAC =1

2× 80

m BAC = 40º

m BAD =1

2m (arc BQD) [Inscribed angle theorem]

30 =1

2 m (arc BQD)

m (arc BQD) = 30 × 2

m (arc BQD) = 60º


6. In the adjoining figure,points B and C lie on tangent tothe circle drawn at point A.Chord AD chord ED.

If m (arc EPF) = 12

m (arc AQD)

and m (arc DRE) = 84º then determine(i) DAC (ii) FDA (iii) FED (iv) BAF (5 marks)

Construction : Draw seg EA.Sol. In DEA,

seg ED seg AD [Given] DEA DAE ......(i) [Isosceles triangle theorem]

But, m DEA =1

2 m (arc AD) .....(ii)

[Inscribed angle theorem]

m DAE =1

2m (arc DRE) .....(iii)

1

2m (arc AQD) =

1

2 m (arc DE) [From (i), (ii) and (iii)]

m (arc AQD) = m (arc DRE)But, m (arc DE) = 84º [Given] m (arc AQD) = 84º

m (arc EPF) =1

2 m(arc AQD) [Given]

m (arc EPF) =1

2 × 84

m (arc EPF) = 42º

A

Q

B

P

C

D

DQ

C

A

B

E

R

PF


SCHOOL SECTION108

A

B

X

C

R1

R2

m (arc AQD) + m (arc DRE) + m (arc EPF) + m (arc AF) = 360º[Measure of a circle is 360º]

84 + 84 + 42 + m (arc AF) = 360 210 + m (arc AF) = 360 – 210 m (arc AF) = 360 – 210 m (arc AF) = 150º

(i) m DAC =1

2 m (arc AQD) [Tangent secant theorem]

m DAC =1

2× 84

m DAC = 42º

(ii) m FDA =1

2 m (arc AF) [Inscribed angle theorem]

m FDA =1

2 × 150

m FDA = 75º

(iii) m FED =1

2 m (arc DAF) [Inscribed angle theorem]

m FED =1

2 [m(arc AQD) + m(arc AF)] [Arc addition property]

m FED =1

2 (84 + 150)

m FED =1

2 × 234

m FED = 117º

(iv) m BAF =1

2 m(arc AF) [Tangent secant theorem]

m BAF =1

2 × 150

m BAF = 75º

Segment of a circle : A secant divides the circular region into twoparts. Each part is called a segment of the circle.

Alternate segment : Each of the two segments formed by the secant ofa circle is called alternate segment in relation with the other.

Angle formed in a segment : An angle inscribed in the arc of a segmentis called an angle formed in that segment.

In the adjoining figure,secant AB divides the circular region intotwo segments R1 and R2.R1 and R2 are alternate segmentsin relation with each other.ACB is inscribed in arc ACB of segment R2.ACB is an angle formed in segment R2.


SCHOOL SECTION 109

ANGLES IN ALTERNATE SEGMENT

AD

B C

X

If a line touches a circle and from the pointof contact a chord is drawn then the angleswhich this chord makes with the given lineare equal respectively to the angles formedin the corresponding alternate segments.In the adjoining figure,

m ABC = 1

2 m (arc AXB) .....(i) [Tangent secant theorem]

m ADB = 1

2 m (arc AXB) ......(ii) [Inscribed angle theorem]

m ABC = m ADB [From (i) and (ii)] ABC ADB

NOTE If a tangent secant angle and an inscribed angle intercept the same arc

then they are congruent.


2. If the chord AB of a circle is parallel to the tangent at C, then provethat AC = BC. (3 marks)

Proof :

Take a point D on the tangent atC as shown in the figure.seg AB || line CD. [Given]

On transversal AC,BAC ACD ........(i) [Converse of alternate angles test]

But, ACD ABC .......(ii) [Angles in alternate segment]In ABC,BAC ABC [From (i) and (ii)]

seg AC seg BC [Converse of Isosceles triangle theorem] AC = BC


3. Suppose AB and AC are equal chords of a circle and a line parallel to thetangent at A intersects the chords at D and E. Prove that AD = AE.

(4 marks)Construction : Draw seg BC.Proof : Take points R and S on the tangent at A

as shown in the figureline DE || line RS [Given]

On transversal AD,EDA DAR [Converse of

alternate angles test] EDA BAR .......(i) [ B - D - A]

BAR BCA .......(ii) [Angles in alternate segment]

A B

D C

B C

D E

R A S•


SCHOOL SECTION110

EDA BCA ......(iii) [From (i) and (ii)]Similarly, we can prove thatDEA CBA ......(iv)In ABC,seg AB seg AC [Given]

BCA CBA ........(v) [Isosceles triangle theorem]In DEA,EDA DEA [From (iii), (iv) and (v)]

seg AD seg AE [Converse of isosceles triangle theorem] AD = AE


4. Suppose ABC is a triangle inscribed in a circle, the bisector of ABCintersects the circle again in D, the tangent at D intersect the line BAand line BC in E and F respectively. Prove that EDA FDC.

Proof :

EDA ABD ........(i)[Angles in alternate segment] FDC CBD ........(ii)

But, ABD CBD .......(iii) [ Ray BD bisects ABC] EDA FDC [From (i), (ii) and (iii)]


5. Suppose two circles are touching at A. Through A two lines are drawnintersecting one circle in P,Q and the other in X. Y. Prove that PQ || XY.

(4 marks)Proof : Two circles are touching at A.

therefore there are two possibilities(a) Two circles touch each other

externally at A.(b) Two circles touch each other

internally at A.(a) If two circles touch each other

externally at A.Proof : Draw a common tangent MN to

the two circles through point A.PQA PAM ......(i) [Angles in alternate segment]

XYA XAN ......(ii)But, PAM XAN .....(iii) [Vertically opposite angles] PQA XYA [From (i), (ii) and (iii)] PQY XYQ [ Q - A - Y] seg PQ || seg XY [By alternate angles test]

B

A C

DE F

• •

P

Q

M

A

N

Y

X


SCHOOL SECTION 111

(b) If two circles touch each other internally at A.

Construction : Draw a common tangent MN to the two circles through point A.

Proof : PQA PAM .......(i) [Angles inXYA XAM alternate segment]

XYA PAM .......(ii) [ P - X - A] PQA XYA [From (i) and (ii)] seg PQ || seg XY [By corresponding angles test]


8. Two circles intersect eachother at A and B. Let DC be acommon tangent touching thecircle C and D.Prove that CAD +CBD = 180º. (3 marks)

Proof : m BAC = m BCD ......(i)[Angles in alternate segments]

m BAD = m BDC ......(ii)In BCD,m BCD + m BDC + m CBD = 180º [Sum of the measures of

angles of a triangle is 180º] m BAC + m BAD + m CBD = 180º [From (i) and (ii)] m CAD + m CBD = 180º [Angle addition property]


9. Let the circles with centre P andQ touch each other at point A.Let the extended chord AB intersectthe circle with centre P at point Eand the chord BC touches the circlewith centre P at the point D.Then provethat ray AD is an angle bisector of the CAE. (4 marks)

Proof :In MAD,MA = MD [The lengths of two tangent


m MAD = m MDA [Isosceles triangle theorem]Let,m MAD = m MDA = xº ........(i)CAM ABC [Angles in alternate segments]Let,m CAM = m ABC = yº ........(ii)m CAD = m CAM + m MAD [Angles Addition property]m CAD = (x + y)º ......(iii) [From (i) and (ii)]

P

Q

X

Y

M

N

A

A

B

C D

PE

AQ

B

MC

D


SCHOOL SECTION112

DAE is an exterior angle of ADB m DAE = m ADB + mABD [Remote Interior angles theorem] m DAE = m ADM + m ABC [D - M - C - B] mDAE = (x + y)º ......(iv) [From (i) and (ii)] m CAD = mDAE [From (iii) and (iv)] ray AD is an angle bisector of CAE.


7. Let M be a point of contact of two internally touching circles. Let lineAMB be their common tangent. The chord CD of the bigger circle touchesthe smaller circle at point N and chord CM and chord DM of biggercircle intersect smaller circle at the points P and R respectively. Provethat CMN DMN. (4 marks)

Construction : Draw seg NR.Proof : CMA CDM [Angles in alternate segments]

Let,m CMA = mCDM = xº ........(i)NMA NRM [Angles in alternate segments]Let,mNMA NRM = yº .......(ii)m NMC = m NMA – m CMA [Angle Addition property]

m NMC = (y – x)º ......(iii) [From (i) and (ii)]NMR DNR .......(iv) [Angles in alternate segment]NRM is an exterior angle of NDR

m NRM = m NDR + mDNR [Remote interior angles] m NRM = CDM + m DNR [ C - N - D and D - R - M] y = x + mDNR [From (i) and (ii)] m DNR = (y – x)º ......(v) m NMR = (y – x)º [From (iv) and (v)] m NMD= (y – x)º ......(vi) [D - R - M ] CMN DMN [From (iii) and (vi)]

ALTERNATIVE METHOD :CMA CDM [Angles

in alternate segments]Let,m CMA = m CDM = xº .....(i)

m NMA = 1

2m (arc NM) .....(ii)

m CNM = 1

2m (arc NM) .....(iii)

m NMA = m CNM [From (ii) and (iii)]Let,m NMA = m CNM = yº ......(iv)m CMN = m NMA – m CMA [Angle addition property]

C DN

P R

MA B

C DN

P R

MA B


SCHOOL SECTION 113

B

A

DC

O

B

A

O

D

C

T

B

P

A

m CMN = (y – x)º ......(v) [From (i) and (iv)]CNM is an exterior angle of NMD,

m CNM = m NDM + m DMN [Remote interior angles theorem] y = x + m DMN [From (i), (vi) and C - N - D] m DMN = (y – x)º ......(vi) CMN DMN [From (v) and (vi)]

THEOREMIf two secants of a circle intersect inside or outside the circle thenthe area of the rectangle formed by the two line segmentscorresponding to one secant is equal in area to the rectangle formedby the two line segments corresponding to the other.

In the adjoining figure,chords AB and CD intersect eachother at point P inside the circle. OA × OB = OC × OD

In the adjoining figure,chords AB and CD intersect eachother at point P outside the circle.

OA × OB = OC × OD

THEOREM

Statement : If a secant and a tangent of a circle intersect in a point outsidethe circle then the area of the rectangle formed by the two line segmentscorresponding to the secant is equal to the area of the square formed by linesegment corresponding to the tangent. (2 marks)

Given :(i) line PAB is a secant intersecting the circle at points A and B.(ii) line PT is a tangent to the circle at point T.

To Prove: PA × PB = PT² Construction: Draw seg BT and seg AT. Proof :In PTA and PBT,

TPA BPT [Common angle]PTA PBT [Angles in alternate segment]

PTA ~ PBT [By AA test of similarity]

PT

PB =

PA

PT[Corresponding sides of similar triangles]

PA × PB = PT²


3. In the adjoining figure,a tangent segment PA touchinga circle in A and a secant PBCare shown. If AP = 15 and BP = 10,find BC. (2 marks)

Sol. Line PBC is a secant intersectingthe circle at points B and C andline PA is a tangent to the circle at point A.

C

B

PA


SCHOOL SECTION114

A

BM N

C

CP × BP = AP² [Tangent secant property] CP × 10 = (15)² CP × 10 = 225

CP =225

10 CP = 22.5 units

CP = BC + BP [ C - B - P] 22.5 = BC + 10 BC = 22.5 – 10

BC = 12.5 units


2. ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD is drawnwhich intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm then findthe values of BD and BX. (2 marks)

Sol. In ABDm BAD= 90º [Angle of a rectangle]

BD2 = AB2 + AD2 [By pythagoras theorem] BD2 = 122 + 92 [Given] BD2 = 144 + 81 BD2 = 225

BD = 15 cm [Taking square roots]

m BAD = 90º [Angle of a rectangle] line BA is a tangent to the circle at point A

[A line perpendicular to the radius at itsouter end is a tangent to the circle]

Line AB is a tangent and line BXD is a secant intersecting at points Xand D

AB2 = BX . BD [Tangent secant property] 122 = BX . 15 144 = BX . 15

BX =144

15

BX = 9.6 cm

EXERCISE - 2.5 (TEXT BOOK PAGE NO. 75)5. In the adjoining figure,

two circles intersect each otherin two points A and B. Seg AB isthe chord of both circles. Point Cis the exterior point of both thecircles on the line AB. From thepoint C tangents are drawn to thecircles touches at M and N.Prove that CM = CN. (2 marks)

Proof : Line CBA is a secant intersecting the circle at points Band A and line CM is a tangent to the circle at point M.

CM² = CB × CA .......(i) [Tangent secant property]Line CBA is a secant intersecting the circle at points B and A andline CN is a tangent to the circle at point N.

CN² = CB × CA ......(ii) [Tangent secant property] CM² = CN² [From (i) and (ii)] CM = CN [Taking square roots]

A D

CB

X

9

12


SCHOOL SECTION 115


6. In the adjoining figure, given twoconcentric circles of radii 5 and 3,find the length of a chord of largercircle which touches the smaller one.If BD = 5, find BC. (3 marks)

Construction : Let the centre of the circle be OLet seg AB touch the smaller circle at E.

Sol. For the smaller circleLine BE is a tangent to the circle at E and line CD is a secantintersecting the circle at points C and D.

BE2 = BC × BD ......(i) [Tangent secant property]In OEB,m OEB = 90º ......(ii) [Radius is perpendicular to tangent]

OB2 = OE2 + BE2 [By Pythagoras theorem] 52 = 32 + BE2 [Given] 25 = 9 + BE2

25 – 9 = BE2

BE2 = 16 BE = 4 units .....(iii) [Taking square roots]

For the larger circle,seg OE chord AB [From (ii)]

AB = 2 × BE [Perpendicular from the centre of thecircle to the chord bisects the chord]

AB = 2 × 4 AB = 8 units (4)2 = BC × 5 [From (i) and (iii)] 16 = BC × 5

BC =16

5

BC = 3.2 units


9. As shown in the adjoining figure, two circlesintersect each other in points A and B.Two tangents touch these circles inpoints P, Q and R, S as shown. Line ABintersects seg PQ in C and seg Rs in D.Show that C and D are midpoints ofseg PQ and RS respectively. (2 marks)

Proof : line PC is tangent to the circle at point P and line CAB is a secant. PC2 = CA × CB .......(i) [Tangent secant property]

Line QC is a tangent to the circle at point C and line CAB is a secant. QC2 = CA × CB ......(ii) [Tangent secant property] PC2 = QC2 [From (i) and (ii)] PC = QC [Taking square roots] C is the midpoint of seg PQ.

Similarly, we can proveD is the midpoint of seg RS.

A

D

O

EC

B

R

P QC

D S

A

B


SCHOOL SECTION116

CD

A B

xCD

A B

x


4. Point M, in the interior of the circle, is the point of intersection of twochords AB and CD of the same circle. Show that CM × BD = BM × AC.

(2 marks)Proof :

CAB BDC ........(i) [Angles inscribedin the same arc are congruent]

In CAM and BDM,CAM BDM [From (i), A - M - B and D - M - C]AMC DMB [Vertically opposite angles]

CAM ~ BDM [By AA test of similarity]

CM

BM =

AC

BD[c.s.s.t.]

CM × BD = BM × AC

Concyclic Points :Given points lying on the same circle arecalled concyclic points.In the adjoining figure,points A,B,C and D are concyclicas all the points lie on the same circle.

NOTE (i) Two distinct points are always concyclic. (ii) Three points in a plane are concyclic if and only if they are non-collinear.

If A, B, C, D are four points such that thepoints C, D are on the same side of the linesegment AB and line segment AB subtendsequal angles at C and D then A, B, C and Dare concyclic.


23. In ABC, seg AB seg AC and P is any point on AC. Through C, a line isdrawn to intersect BP produced in Q such that ABQ ACQ. Prove

that AQC = 90º + 12BAC. (2 marks)

Proof :

In ABC,seg AB seg AC [Given]

ABC ACB ......(i) [Isosceles triangle theorem]ABQ ACQ [Given]

seg AQ subtends congruent angles at points B and C which are on thesame side of line AQ.

Points A, B, C, Q are concyclic.

A

B C

Q

P

AD

M

C

B


SCHOOL SECTION 117

ABCQ is cyclic [By definition] m AQC + m ABC = 180º [ Opposite angles of a cyclic

quadrilateral are supplementary] m AQC = 180 – m ABC ......(ii)

In ABC,m ABC + m ACB + m BAC = 180º [ Sum of the measure of angles

of a triangle is 180º] m ABC + m ABC + m BAC = 180º [From (i)] 2m ABC + m BAC = 180 2m ABC = 180 – m BAC

m ABC = 90 – 1

2 m BAC .....(iii) [Dividing throughout by 2]

m AQC = 180 – (90 – 1

2 m BAC) [From (ii) and (iii)]

m AQC = 180 – 90 + 1

2m BAC

m AQC = 90º + 12

m BAC


22. ABC is inscribed in a circle.Point P lies on a circumscribed circle of atriangle such that through point P, PN,PM and PL are perpendicular on sides oftriangle (possibly by increasing sides).Prove that point N, M and L are collinear. (5 marks)

Construction : Draw seg NM, seg ML and seg AP.Proof : m PNA = 90º ......(i) [Given]

m PMA = 90º ......(ii)Adding (i) and (ii),m PNA + m PMA = 90º + 90º

m PNA + PMA = 180º PNAM is cyclic [If opposite angles of a quadrilateral

are supplementary then thequadrilateral is cyclic]

Point P, N, A and M are concyclic. PAN PMN .....(iii) [Angles inscribed in same arc are congruent]

PABC is cyclic [By definition] PAN PCB [Exterior angles of a cyclic quadrilateral

is equal to its interior opposite angle] PAN PCL .....(iv) [C - L - B] PMN PCL ......(v) [From (iii) and (iv)]

PMC PLC [ each is 90º] seg PC subtends congruent angles at points M and L lying on the same

side of line PC Points P, M, L and C are concyclic PMLC is cyclic [By definition] m PCL + mPML = 180º .....(vi)[Opposite angles of cyclic quadrilateral

are supplementary] m PMN + m PML = 180º [From (v) and (vi)]

Also, PMN and PML are adjacent angles PMN and PNL form a linear pair [Converse of linear pair axiom] ray MN and ray ML are opposite rays Points N, M and L are collinear.

NA

P

M

LC

B


SCHOOL SECTION118

HOTS PROBLEM(Problems for developing Higher Order Thinking Skill)

9. In the adjoining figure,DE intersects the sides of ABCin point P and Q such thatArc AD = Arc DB and Arc AE = Arc EC.Show that PQC BPQ. (5 marks)

Construction : Draw seg DC and seg BFProof : m (arc AD) = m (arc BD) ......(i) [Given]

m ACD = 1

2m (arc AD) ......(ii) [Inscribed angle theorem]

m BED = 1

2m (arc BD) .....(iii)

m ACD = m BED [From (i), (ii) and (iii)] ACD BED QCD BEP ......(iv) [ A - Q - C and D - P - E]

m (arc AE) = m (arc CE) .......(v) [Given]

m ABE = 1

2 m (arc AE) ....(vi) [Inscribed angle theorem]

m CDE = 1

2 m (arc CE) .....(vii)

m ABE = m CDE [From (v), (vi) and (vii)] ABE CDE PBE CDQ ......(viii) [ A - P - B and D - Q - E]

In BEP and DCQ,BEP QCD [From (iv)]PBE CDQ [From (viii)]

BEP ~ DCQ [By AA test of similarity] BPE DQC [c.a.s.t.] BPQ PQC [ P - Q - E and D - P - Q]

10. In ABC, A is an obtuse angle.P is the circumcentre of ABC.Prove that PBC = A – 90º.

(5 marks)

Construction : Extend seg BP to intersect thecircle at point D, B - P - D anddraw seg AD.

Proof : m BAD = 90º .......(i) [Angle subtended by a semicircle]DBC DAC [Angles inscribed in same arc are

congruent]i.e. PBC DAC ......(ii) [B - P - D]

BAC = BAD + DAC [Angle addition property] A = 90º + PBC [From (i) and (ii)] A – 90 = PBC PBC = A – 90º

A

PQ

D

E

B

C

A

PB

D

C


SCHOOL SECTION 119

11. In the adjoining figure,BC is a diameter of the circlewith centre M. PA is a tangentat A from P which is a point online BC. AD BC.Prove that DP2 = BP × CP – BD × CD.

(5 marks)Construction : Draw seg AB and seg ACProof : In ADP,

m ADP = 90º [Given] AP2 = AD2 + DP2 [By Pythagoras theorem] DP2 = AP2 – AD2 ......(i)

Line AP is a tangent to the circle at point A and line PCB is a secant tothe circle at points C and B

AP2 = BP × CP ......(ii)In BAC,m BAC = 90º [Angle subtended by semicircle]seg AD hypotenuse BC

AD2 = BD × CD ......(iii) [Property of geometric mean] DP2 = BP × CP – BD × CD [From (i), (ii) and (iii)]

12. ABC is an equilateral triangle.Bisector of B intersectscircumcircle of ABC in point P.Prove that CQ = CA.

(5 marks)

Proof : ABC is an equilateral triangle [Given] m ABC = m BAC = m ACB = 60º ......(i)

[Angle of an equilateral triangle]

m CBP = 1

2m ABC [ ray BP bisects ABC]

m CBP = 1

2 × 60 [From (i)]

m CBP = 30º .....(ii)CBP CAP ....(iii) [Angles inscribed in the same

are congruent] m CAP = 30º [From (ii) and (iii)] m CAQ = 30º .....(iv) [ A - P - Q]

ACB is an exteror angle of CQA m ACB = m CQA + m CAQ [Remote interior angle theorem] 60 = m CQA + 30 [From (i) and (iv)] m CQA = 60 – 30 m CQA = 30º .....(v)

In CQACAQ CQA [From (iv) and (v)]

seg CQ seg CA [Converse of Isosceles triangle theorem] CQ = CA

A

B M D C P

B

C A

P

Q


SCHOOL SECTION120

13. Two circles with centre O andP intersects each other in poitn Cand D. Chord AB of the circle withcentre O touches the circle withcentre P in point E.Prove that ADE + BCE = 180º. (4 marks)

Construction : Draw seg DC.Proof : ABCD is cyclic [By definition]

m CBE = m ADC ......(i) [Exterior angle property of cyclicquadrilateral]

m CDE = m CEB ......(ii) [Angles in alternate segment]In BCE,m CBE + m CEB + m BCE = 180º [Sum of the measures of the

angles of a triangle is 180º] m ADC + m CDE + m BCE = 180º [From (i) and (ii)] m ADE + m BCE = 180º [Angle addtion property]

58. Show that the radius of incircle of right angle triangle is equal to thedifference of half of the perimeter and the hypotenuse. (5 marks)

Given : (i)In ABC, m ABC = 90º(ii) Circle with centre O touches the

sides AB, BC and AC at point L, Mand N respectively.

To prove : r = 1

2 (AB + BC + AC) – AC

Proof : Let the radius of the incircle be r OL = OM = ON = r [Radii of the same circle]

AL = AN = x .....(i) [Lengths of the two tangentsBL = BM = y .....(ii) segment from an external point toCM = CN = z ....(iii) a circle are equal]AB + BC + AC = AL + BL + BM + CM + AN + CN

[A - L - B, B - M - C and A - N - C]AB + BC + AC = x + y + y + z + x + z [From (i), (ii) and (iii)]

AB + BC + AC = 2x + 2y + 2z AB + BC + AC = 2 (x + y + z)

1

2 (AB + BC + AC) = x + y + z

1

2 (AB + BC + AC) = (x + z) + y

1

2 (AB + BC + AC) = AC + y [From (i), (iii) and A - N - C]

1

2(AB + BC + AC) – AC = y ......(iv)

In LBMO,m LBM = 90º [Given]m BLO = m BMO = 90º [Radius is perpendicular to tangent]

m LOM = 90º [Remaining angle] LBMO is a rectangle [By definition] LB = OM [Opposite sides of an rectangle] LB = OM = y = r ......(v)

12

(AB + BC + AC) – AC = r [From (iv) and (v)]

O

D

P

CA B E

A

B C

N

L

M

xx

z

zy

yO


SCHOOL SECTION 121

59. If two circles are internally touching at point P. A line intersect thosetwo circles in point A, B, C, D respectively then prove that APB CPD.

(5 marks)

Construction : Draw a common tangent MN at point P.Proof : APM ADP [Angles in alternate segment]

Let,m APM = m ADP = x ......(i)BPM BCP [Angles in alternate segment]Let,m BPM = m BCP = y .....(ii)m APB = m BPM – m APM [Angle addition proerty]

m APB = (y – x) .....(iii) [From (i) and (ii)]BCP is an exterior angle of CPD,

m BCP = m CPD + m CDP [Remote interior angles theorem] y = m CPD + x [From (i) and A - C - D] m CPD = (y – x) ......(iv) m APB = m CPD [From (iii) and (iv)] APB CPD

MCQ’s1. P is the centre of the circle. T is a point on the circle and TB is a tangent.

PBT = 30º and radius is 10 cm. What is length of TB ?(a) 20 cm (b) 20 3 cm

(c) 10 cm (d) 10 3 cm

2. TP and TQ are tangents to the circle with centre O. If PTQ = 80º thenwhat is the measure of POT ?(a) 100º (b) 80º(c) 50º (d) 40º

3. Two circles with centres P and Q touch externally at T. AB is a commontangent to the circles. What is the measure of ATP ?(a) 30º (b) 45º(c) 60º (d) 90º

4. Circles with centres A and B and radius 10 cm and 5 cm touch each otherinternally at point C. P is any point on AC such that AP = 7 cm. How manytangents can be drawn to the circles from point P ?(a) 0 (b) 1(c) 2 (d) 1 or 2

C

B

P

M

N

A

D


SCHOOL SECTION122

5. P is the centre of the circle. AB is a tangent to the circle at A. PAB is anisosceles triangle. What is the measure of APB ?(a) 30º (b) 45º(c) 60º (d) 90º

6. A circle is inscribed in ABC in which BC = 6 cm, AB = 8 cm and B = 90º.What is the radius of the circle ?(a) 4 cm (b) 3 cm(c) 2 cm (d) 9 cm

7. ABC is inscirbed in a circle AB = 15 cm, AC = 8 cm, A = 90º. What is theradius of the circle ?(a) 8.5 cm (b) 8 cm(c) 7.5 cm (d) 4 cm

8. Rectangle ABCD is inscribed in a circle with centre P and diameter 41 cm. Ifone of the sides of the rectangle is 9 cm then what is the perimeter ?(a) 49 cm (b) 50 cm(c) 98 cm (d) 100 cm

9. ABCD is cyclic quadrilateral such that 2 A = 3 C. What is the measureof C ?(a) 108º (b) 72º(c) 100º (d) 80º

10. In the adjoining figure,TA and TB are tangents from T.CD is also a tangent at P.It TB = 12 cm. Find perimeter of TCD.(a) 36 cm (b) 18 cm(c) 20 cm (d) 24 cm

11. seg AT is a tangent at T for a circle with centre O. TOB is the diameter. IfAOB = 130º then what is measure of OAT ?(a) 40º (b) 50º(c) 65º (d) 90º

12. An angle substended by a semicircle at a point outside the circle is.................. angle.(a) right (b) an acute(c) an obtuse (d) 180º

13. A cyclic parallelogram is ................ .(a) square (b) rhombus(c) rectangle (d) kite

14. If two secants AB and CD of a circle intersect in point ‘P’ then

(a)AP CP

BP PD (b) AP = BP

(c) CP = CD (d) AP × BP = CP × DP

15. If circles with centres A, B and C with radius 5 cm, 6 cm and 7 cm toucheach other externally. The perimeter of ABC is ............. cm.(a) 18 (b) 23(c) 36 (d) 40

A

B

P

D

C

T


SCHOOL SECTION 123

16. Two concentric circles are of radii 5 cm and 3 cm. The length of the chordof the larger circle which touches the smaller circle will be .................. .(a) 8 (b) 4(c) 6 (d) 10

17. A chord divides a circle into 2 arcs measuring 2x and 7x. The measure ofthe minor arc is .............. .(a) 20 (b) 40(c) 80 (d) 140

18. seg AB and seg AD are chordsof the circle and AC is a tangent.If m (arc APB) = 70º, m (arc BQD) = 16º,then m CAD = ............. .

(a) 35 (b) 43(c) 32 (d) 27

19. Two circles with centres P and Q having diameter 25 cm and 15 cmrespectively touch each other externally at A, then the distance between Pand Q is ............. .(a) 40 cm (b) 10 cm(c) 20 cm (d) 45 cm

20. seg AB is such that APB AQB on the same side of seg AB. Hence,points A, B, C, D are ............. .(a) collinear (b) coinciding(c) concyclic (d) non-concylic

21. MT is a tangent and MSB is a secant MS = 4, MB = 16, hence MT = ?(a) 8 (b) 10(c) 12 (d) 30

22. In the adjoining figure,seg PT is a tangent and seg PB is secant.PTA ~ ............. .

(a) PAT (b) BPT(c) PBT (d) TAB

23. The diameter of two circles touching each other externally are 27 cm and13 cm. Find the distance between their centre ............ .(a) 20 cm (b) 21 cm(c) 19 cm (d) 12 cm

24. In a circle with centre P. Line QR is tangent at R. PQ = 13 cm, QR = 12 cm, thenradius PR = ......... cm.(a) 25 (b) 5(c) 10 (d) 1

25. If TP and TQ are the two tangents to a circle with centre O. So thatPOQ = 100º, then PTQ = ?(a) 60 (b) 80(c) 50 (d) 120

DQ

B

P

A C

•

•

A

T

BP


SCHOOL SECTION124

: ANSWERS :

1. (d) 10 3 cm 2. (a) 100º

3. (d) 90º 4. (a) 05. (b) 45º 6. (c) 12 cm7. (a) 8.5 cm 8. (c) 98 cm9. (b) 72º 10. (d) 24 cm11. (a) 40º 12. (b) an acute13. (c) rectangle 14. (d) AP × BP = CP × DP15. (c) 36 16. (a) 817. (c) 80 18. (b) 4319. (c) 20 cm 20. (c) concyclic21. (a) 8 22. (b) PBT23. (a) 20 cm 24. (b) 525. (b) 80

1 Mark Sums1. In the adjoining figure,

if m ABC = 55º, thenwhat is m AEB.

Sol. AEB ABC [Angles in alternate segments]But, ABC = 55º [Given]

AEB = 55º

2. What is the relation betweenABC and ADC of cyclic ABCD ?

Sol.

ABCD is cyclic [Given] m ABC + m ADC = 180º [Opposite angles of quadrilateral are

supplementary] ABC and ADC are supplementary.

3. If seg AB is the diameter of the circle,then find AB, if OB = 3 and OA = 4.

Sol.

seg AB is the diameter of the circle. m AOB = 90º ......(i) [Angle subtended by a semicircle]

In AOB,m AOB = 90º [From (i)]AB2 = OA2 + OB2 [By Pythagoras theorem]

AB2 = 42 + 32

AB2 = 16 + 9 AB2 = 25

AB = 5 units [Taking square roots]

E A

B C•

AB

C

D

A

B

O


SCHOOL SECTION 125

4. O is the centre of the circle.If m ABC = 80º, the findm (arc AC) and m (arc ABC).

Sol. m ABC = 1

2 m (arc AC) [By Inscribed angle theorem]

80º = 1

2 m (arc AC)

m (arc AC) = 160º

m (arc ABC) = 360º – m (arc AC)= 360 – 160

m (arc ABC) = 200º

5. If PB = 3, PD = 4, PA = 6, find PC.Sol. Chords AB and CD intersect each

other at point P outside the circle. PA × PB = PC × PD 6 × 3 = PC × 4

PC = 6 3

4

PC = 9

2 PC = 4.5 units

6. What is the relation between ABEand ADC for cyclic ABCD ?

Sol.

ABE is an exterior angle of cyclic ABCD ABE ADC [An exterior angle of a cyclic quadrilateral is

congruent to its interior opposite angle]

7. P is the centre of the circle andline AB is a tangent at T. Radiusof the circle is 5 cm. Find thedistance of P from line AB.

Sol.seg PT line AB [Radius is perpendicular to the radius]PT = 5 cm [Given]

Distance of P from line AB is 5 cm

8. Lines PM and PN are tangents tothe circle with centre O.If PM = 7 cm, find PN.

Sol.

PM = PN [Length of the two tangent segments from anexternal point to a circle are equal]

But, PM = 7 cm [Given]

PN = 7 cm

B

O

A

C

AB

D

P

C

A

BC

D

E•

P

A T B

M

O

N

P


SCHOOL SECTION126

9. If m (arc PNQ) = 140º,find m PQR.

Sol.

m PQR = 1

2 m (arc PNQ) [Tangent-secant theorem]

m PQR = 1

2 × 140º

m PQR = 70º

10. Line AB is a tangent and line BCDis a secant. If AB = 6 units, BC = 4 units,find BD.

Sol. Line BCD is a secant intersectingthe circle at points C and D andline BA is a tangent at A

AB2 = BC × BD 62 = 4 × BD 36 = 4 × BD

BD = 36

4

BD = 9 units

11. Two circles with centres P and Q having diameter 25 cm and 15 cmrespectively touch each other externally at A. Find the distance betweenP and Q.

Sol. Diameter of bigger circle = 25 cm Its radius (r1) = 12.5 cm

Diameter of smaller circle = 15 cm Its radius (r2) = 7.5 cm

The two circles with centres P and Q touch each other externally at A. Distance between P and Q = r1 + r2

= 12.5 + 7.5= 20 cm

The distance between P and Q is 20 cm.

12. In the adjoining figure,chords AB and CD intersect at E.If DE = 6, BE = 3 and CE = 4, thenfind AE.

Sol. Chords AB and CD intersect eachother at point E inside the circle

AE × BE = CE × DE AE × 3 = 4 × 6

AE = 4 6

3

AE = 8 units

P

N

Q

R

A

B CD

A

BD

C

E


SCHOOL SECTION 127

13. In the adjoining figure,m ABC = 57º. M is the centreof the circle and line BC is a tangent.seg BP is the diameter find,(a) m (arc BQA) (b) m ABP.

Sol.

(a) m ABC = 1

2 m (arc BQA) [Tangent-secant theorem]

57º = 1

2 m (arc BQA)

m (arc BQA) = 114º

(b) seg PB is the diameter m (arc PAB) = 180º [Measures of a semicircle] m (arc PA) + m (arc BQA) = m (arc PAB)

[Arc addition property] m (arc PA) + 114º = 180º m (arc PA) = 180º – 114º m (arc PA) = 66º

m ABP = 1

2 m (arc PA) [Inscribed angle theorem]

m ABP = 1

2 × 66º

m ABP = 33º

14. O is the centre of a circle. TA is a tangent to the circle at point T. Whatis measure of OTA ?

Sol. Line TA is a tangent to the circle at point T [Given]

m OTA = 90º [Radius is perpendicular to the tangent]

15. Radius of the circle is 4 cm. What is the length of a chord of the circle ?Sol. Radius of the circle is 4 cm

Diameter of the circle = 4 × 2 = 8 cmDiameter is the biggest chord of the circle

The length of the chord of the circle is 8 cm or less than 8 cm.

16. O is the centre of the circle. AB is the diameter. P is any point on thecircle other than A and B. What is m APB ?

Sol. seg AB is the diameter of the circle and P is the point on the circle

m APB = 90º [Angle subtended by a semicircle]

17. What is the relation between the diameter and the radius of a circle ?Sol. Diameter of a circle is twice its radius.

18. In the adjoining figure,line PQ and line PR are tangentsto the circle. So what ism QPR + m QOR ?

Sol.

m OQP = 90º ......(i)[Radius is perpendicular to the tangent]m ORP = 90º ......(ii)

A

B C

M

P

Q

Q

O

R

P


SCHOOL SECTION128

In OQPR,m QPR + m QOR + m OQP + m ORP = 360º

[Sum of the measures of angles of aquadrilateral is 360º]

m QPR + m QOR + 90º + 90º = 360º [From (i) and (ii)] m QPR + m QOR = 360º – 180º

m QPR + m QOR = 180º

19. O is the centre of the circle. AB is the longest chord of the circle. IfAB = 8.6 cm, what is the radius of the circle ?

Sol. AB is the longest chord of the circle [Given]But, diameter is the longest chord of the circle

seg AB is the diameter of the circleDiameter = 8.6 cm

Radius = 4.3 cm

20. In the adjoining figure,if m APB = 30º, then m AOB ?

Sol.

m APB =1

2 m (arc AB) [Inscribed angle theorem]

30º =1

2 m (arc AB)

m (arc AB) = 60ºm AOB = m (arc AB) [Definition of measure of minor arc]

m AOB = 60º

BO

A

P

MAHESH TUTORIALS PVT. LTD.

MT EDUCARE PVT. LTD. GEOMETRY

7

Q.I. Solve the following : (4)(i) In the adjoining figure,

Q is the centre of circle and PMand PN are tangent segments tothe circle. If MPN = 40º circle,find MQN.

(ii) If two circles with radii 8 and 3 respectively touch internally thenshow that the distance between their centers is equal to the differenceof their radii, find that distance.

Q.II. Attempt the following : (9)(i) In the adjoining figure,

are four tangents to a circle at thepoints A,B, C and D. These four tangentsform a parallelogram PQRS.If PB = 5 and BQ = 3 then find PS.

(ii) In the adjoining figure,line AB is tangent to both thecircles touching at A and B.OA = 29, BP = 18, OP = 61then find AB.

(iii) Three congruent circles withcentres A, B and C and withradius 5 cm each, touch eachother in points D, E, G as shown in(a) What is the perimeter of ABC ?(b) What is the length of side DE of DEF ?

CHAPTER 2 : Circle

SET - A

S.S.C.

GEOMETRY

Marks : 30

Duration : 1 hr. 15 min.

P

N

Q

M

40º

BP Q

AC

SD

R

5 3

O

A B

P

18

61

M29

A B

C

D

EF

GEOMETRY MT EDUCARE PVT. LTD.

8 MAHESH TUTORIALS PVT. LTD.

Q.III. Solve the following : (12)(i) In the adjoining figure,

point A is a common point of contactof two externally touching circlesand line l is a common tangentto both circles touching at B and C.Line m is another common tangentat A and it intersects BC at D.Prove that (a) BAC = 90º(b) Point D is the midpoint of seg BC.

(ii) In the adjoining figure,points P, B and Q arepoints of contact of the respectivetangents. line QA is parallel to line PC.If QA = 7.2 cm, PC = 5 cm, find theradius of the circle.

(iii) In a right angled triangle ABC, ACB = 90º acircle is inscribed in the triangle with radius r.a, b, c are the lengths of the sides BC, AC andAB respectively. Prove that 2r = a + b – c.other circle at A. Prove that thepoints D, T and A are collinear.

Q.IV. Solve the following : (5)(i) In the adjoining figure,

points P and Q are the centers ofthe circles. Radius QN = 3, PQ = 9.M is the point of contact of the circles.Line ND is tangent to the larger circle.Point C lies on the smaller circle.Determine NC, ND and CD.

B

A

m

l

CD

CP

OB

Q A

A

O

Q

P

C R B

CD

Q PN M

Best of Luck



9


if m (arc APC) = 60ºand m BAC = 80ºFind (a) ABC (b) m (arc BQC)

(ii) In the adjoining figure,m (arc XAZ) = m (arx YBW).Prove that XY || ZW

Q.II. Solve the following : (9)(i) Secants containing chords RS

and PQ of a circle intersects eachother in point A in the exterior ofa circle, as shown in figure Ifm (arc PCR) = 26º andm (arc QDS) = 48ºthen find(a) AQR (b) SPQ, (c) RAQ

(ii) Two circles intersect each otherin points A and B. Secants throughA and B intersects circles in C, D andM, N. Prove that CM || DN.

CHAPTER 2 : Circle

SET - B

S.S.C.

GEOMETRY

Marks : 30


B

A

C80º

•Q

•P

X Y

A B

Z W

D•

S

Q

C

R

P

A

C AD

M B N



(iii) Find the radius of the circle passing through the vertices of a rightangled triangle when lengths of perpendicular sides are 6 and 8.

Q.III. Solve the following : (12)(i) ABCD is cyclic quadrilateral, lines AB and DC

intersect in the point F and lines AD and BCintersect in the point E. Show that thecircumcircle of BCF and CDE intersectin a point on the line EF.

(ii) In a cyclic quadrilateral ABCD,the bisectors of opposite angles Aand C meet the circle at P and Qrespectively. Prove that PQ is adiameter of the circle.

(iii) In DABC, midpoints of sides AB,AC and BC are P, Q and R respectively.AS BC. Prove that PQRS is acyclic quadrilateral.

Q.IV. Solve the following : (5)(i) Prove that the quadrilateral formed by the angle bisectors of a cyclic

quadrilateral is also cyclic.

FG

C

E

B D

A

D P

Q

A

C

B

••

××

A

P Q

BS R

C

Best of Luck



11


seg AB and seg AD are chordsof the circle. C be a point ontangent to the circle at point A.If m (arc APB) = 80º and BAD = 30º,then find (i) BAC (ii) m (arc BQD)

(ii) In the adjoining figure,a tangent segment PA touchinga circle in A and a secant PBCare shown. If AP = 15 and BP = 10,find BC.

Q.II. Solve the following : (9)(i) In the adjoining figure,

two circles intersect each otherin two points A and B. Seg AB isthe chord of both circles. Point Cis the exterior point of both thecircles on the line AB. From thepoint C tangents are drawn to thecircles touches at M and N.Prove that CM = CN.

(ii) Two circles intersects eachother at A and B. Let DC be acommon tangent touching thecircle C and D.Prove that CAD +CBD = 180º.

CHAPTER 2 : Circle

SET - C

S.S.C.

GEOMETRY

Marks : 30


A

Q

B

P

C

D

C

B

PA

A

BM N

C

A

B

C D



(iii) In the adjoining figure, given twoconcentric circles of radii 5 and 3,find the length of a chord of largercircle which touches the smaller one.If BD = 5, find BC.

Q.III. Solve the following : (12)(i) Suppose AB and AC are equal chords of a circle and a line parallel to

the tangent at A intersects the chords at D and E. Prove that AD = AE.

(ii) Let M be a point of contact of twointernally touching circles. Let lineAMB be their common tangent.The chord CD of the bigger circletouches the smaller circle at point Nand chord CM and chord DM of biggercircle intersect smaller circle at thepoints P and R respectively.Prove that CMNDMN

(iii) Let the circles with centre P andQ touch each other at point A.Let the extended chord AB intersectthe circle with centre P at point Eand the chord BC touches the circlewith centre P at the point D.Then provethat ray AD is an angle bisector of the CAE.

Q.IV. Solve the following : (5)(i) In the adjoining figure,

points B and C lie on tangent tothe circle drawn at point A.Chord AD chord ED.

If m (arc EPF) = 1

2 m (arc AQD)

and m (arc DRE) = 84º then determine(a) DAC (b) FDA (c) FED (d) BAF

A

D

O

EC

B

C DN

P R

MA B

PE

AQ

B

MC

D

DQ

C

A

B

E

R

PF

Best of Luck

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