2005 b.sc maths project

APPLICATION OF TRANSFORMATION

GEOMETRY TO SOME CLASSICAL

THEOREMS

Mariyam Shahuneeza Naseer

B. Sc. Mathematics Project

i

APPLICATION OF TRANSFORMATION

GEOMETRY TO SOME CLASSICAL

THEOREMS

An assignment submitted in partial fulfilment of the

requirements of the subject “Mathematics Seminar”

by


November 2005

ii

ACKNOWLEDGEMENTS

I am grateful to my supervisor Dr. Kauta, for giving me the opportunity to work in a

very interesting area, and for his support and guidance throughout the preparation of this

report. Thank you very much, Dr. Kauta.

I wish to express my sincere thanks to Dr. Anderson and Dr. Bong for their invaluable

help.

Finally, I wish to thank my father for checking my report before I submitted it to

Universiti Brunei Darussalam (UBD).


November 2005

iii

FOREWORD

The main purpose of preparing the report was to fulfil part of the coursework

requirements for a Universiti Brunei Darussalam subject MA 4238, “Mathematics Seminar.”

The report applies the methods of transformation geometry to some important theorems

of classical geometry.

Transformation geometry is that area of geometry that deals with certain types of

mappings of points and lines, called dilations and dilatations.

In the report that follows I have divided the Classical Theorems into four main divisions.

They are Menelaus’ Theorem, Ceva’s Theorem, Desargues’ Theorem, and Pappus’ Theorem.

Menelaus’ Theorem, which involves a test for the collinearity of three points, and Ceva’s

Theorem, which involves a test for the concurrency of three lines, are frequently called the

Twin Theorems.

Desargues’ Theorem shows a relationship between the concurrency of three lines and the

collinearity of three points, while Pappus’ Theorem shows how two sets of three collinear

points can be used to construct a third set of three collinear points.


November 2005

Table of Contents

iv

TABLE OF CONTENTS Acknowledgements ii

Foreword iii

Table of Contents iv

CHAPTER 1: INTRODUCTION 1

Lemma 1.1 1

Lemma 1.2 4

Lemma 1.3 7

CHAPTER 2: MENELAUS’ THEOREM 8

Theorem 2.1 8

CHAPTER 3: CEVA’S THEOREM 11

Theorem 3.1 11

CHAPTER 4: DESARGUES’ THEOREM 15

Theorem 4.1 15

CHAPTER 5: PAPPUS’ THEOREM 18

Theorem 5.1 18

References 21

APPENDICES 22

Appendix A: Proof of Lemma 1.2 (Case 2 and Case 3) 22

Appendix B: Definitions 28

Appendix C: Theorems That Were Made Use Of 30

CHAPTER 1

INTRODUCTION

The classical theorems you come across in this report will depend mostly on the

following lemma.

It is noteworthy that some of the terms, such as a dilation, a dilatation, a fixed point,

a fixed line, the identity and the symbol δP,r are defined in Appendix B (p.28).

Lemma 1.1:

a. A dilatation with a fixed point off a fixed line is the identity.

b. If A ≠ B and ab ≠ 1, then δB,b δA,a is a dilation with centre on AB .

c. If δR,r δQ,q δP,p = ι and one of the numbers p, q, r is different from 1, then points P, Q,

R, are collinear.

Proof:

(a)

Figure 1.1.1

Dilatation δ fixes point Z as well as the line l.

δ (Z) = Z

δ (l) = l.

Now we draw a line m through Z which intersects the line l at point C.

Figure 1.1.2

Since dilatation δ takes a line to a line which is parallel to the original line, m ║ δ (m).

The point Z is on the line m. Hence, δ (Z) is on the line δ (m).

But δ (Z) = Z.

Application of Transformation Geometry to some Classical Theorems: Introduction

2

Therefore, the point Z lies on the line m and δ (m).

Since the line m is parallel to the line δ (m) and Z is on both lines, we have m = δ (m).

Since the point C is the intersection of lines l and m, δ (C) is the intersection of lines δ (l) and

δ (m).

But δ (l) = l and δ (m) = m. therefore, δ (C) = C.

Hence, dilatation δ fixes the two points C and Z.

By Theorem 1 (Appendix C) there is a unique dilatation which takes points Z to Z and C to C.

This dilatation is the identity, so δ = ι.

(b)

Suppose A ≠ B and ab ≠ 1.

If δB,bδA,a = ι

Then δB,bδA,a (A) = A.

But δA,a (A) = A.

δB,bδA,a (A) = δB,b(A)= A.

BA = bBA

b = 1.

This implies that δB,b = ι.

δB,bδA,a = ι δA,a = δA,a.

This implies that δA,a = ι.

a = 1.

This contradicts our assumption that ab ≠ 1.

Let l = AB .

Look at δB,b δA,a (l):

The point A is on the line l.

δA,a (l)= l.

This implies δB,b δA,a (l) = δB,b (l)

= l, since the point B is on the line l.

δB,b δA,a fixes the line AB .

By Theorem 3 (Appendix C), if A ≠ B and ab ≠ 1, then

δB,b δA,a = δC,ab for some point C.

By part (a), the centre of dilation C should be on the line AB since δB,bδA,a ≠ ι.


3

(c)

If any two of the three points are the same, then the three points will always be collinear as we

can draw a straight line through any two points.

Therefore, we assume P, Q, R are distinct and one of the numbers p, q, r is different from 1.

δR,r δQ,q δP,p = ι

δQ,q δP,p = δ-1

R,r

= δR,r

1 .

Suppose r = 1:

Then, δQ,q δP,p = ι

Therefore δQ,q δP,p (P) = P

Since δP,p (P) = P we have δQ,q (P) = P

QP = qQP

q = 1.

Therefore δQ,q δP,p = ι δP,p.

This implies, δP,p = ι

p = 1.

This contradicts our assumption that one of the numbers p, q, r is different from 1.

Therefore, r ≠ 1.

We have δQ,q δP,p = δR,r

1 .

Let l = PQ .

Look at δQ,q δP,p (l):

The point P is on the line l.

δP,p (l)= l.

This implies δQ,q δP,p (l) = δQ,q (l)

= l, since the point Q is on the line l.

δQ,q δP,p fixes the line PQ .

δR,r

1 fixes the line PQ as δR,r

1 = δQ,q δP,p .

δR,r

1 fixes the point R, since R is the centre of dilation.

So δR,r

1 fixes the point R and the line PQ .


4

But δR,r

1 ≠ ι since r ≠ 1.

Therefore, by part (a), R is on the line PQ .

So, points P, Q, R are collinear.

Lemma 1.2:

Let A, B, C, D, E be distinct points and r and s be non-zero real numbers such that

δA,r (C) = δB,s (D) = E.

Then r = s iff AB ║ CD .

Proof:

Case 1 (0 < r < 1):

Figure 1.2.1

It is given that

δA,r (C) = E and

AE = r AC,

AE / AC = r

δB,s (D) = E

BE = r BD,

BE / BD = s

Suppose r = s:

Then we have

AE / AC = BE / BD

Let x = AE / AC = BE / BD

AE = x AC and also EC = AC – AE

AE / EC = x AC / (AC – AE)

= (x AC / AC) / ([AC / AC] – [AE / AC])

= x / (1 – x).


5

BE = x BD and also ED = BD – BE

BE / ED = x BD / (BD – BE)

= (x BD / BD) / ([BD / BD] – [BE / BD])

= x / (1 – x).

EC = AC – AE and

= AC – xAC

= (1 – x)AC

ED = BD – BE

= BD – xBD

= (1 – x)BD.

Then

AB2 = AE

2 + BE

2 – 2AE BEcosθ

= x2AC

2 + x

2BD

2 – 2x

2AC BDcosθ

= x2 [AB

2 + BD

2 – 2AC BDcosθ]

and

CD2 = EC

2 + ED

2 – 2EC EDcosθ

= (1-x)2AC

2 + (1-x)

2BD

2 – 2(1-x)

2AC BDcosθ

= (1-x)2 [AB

2 + BD

2 – 2AC BDcosθ].

AB2 x

2 [AB

2 + BD

2 – 2AC BDcosθ]

=

CD2 (1-x)

2 [AB

2 + BD

2 – 2AC BDcosθ]

( AB/ CD) = x / (1 – x).

Since the three pairs of corresponding sides are proportional, the triangles

∆ ABE ~ ∆CDE.

This implies that the three pairs of the corresponding angles are equal.

mAEB = mDEC

mBAE = mDCE

mABE = m CDE

Hence, the lines AB and CD are parallel.

Therefore, if r = s, AB ║ CD .

Conversely, suppose the lines AB and CD are parallel.



6

m AEB = m DEC (vertical pair)

m BAE = m DCE (alternate interior angles)

m ABE = m CDE (alternate interior angles)

Hence, triangles ∆ ABE ~ ∆CDE.

Therefore, the corresponding sides must be proportional.

Let a = AE / EC = BE / ED.

Then, we have

AE AE

= = a

EC AC – AE

AE = a (AC – AE)

(1 + a) AE = a AC

AE a

=

EC (1 + a)

and

BE BE = = a

ED BD – BE

BE = a (BD – BE)

(1 + a) BE = a BD

BE a

= BD (1 + a)

Therefore, AE / AC = a / (1 + a) = BE / BD.

This implies r = AE / AC = BE / BD = s.

Therefore, if AB ║ CD , then r = s.

Case 2 (r > 1): Case 3 (r < 0):

Figure 1.2.2 Figure 1.2.3


7

A similar approach as in Case 1 can be used to prove Case 2 and Case 3. For the complete

proof of Case 2 and Case 3 refer to Appendix A (p. 22)

Lemma 1.3:

If distinct lines AB and PQ intersect at point L, then

δA,a(P) = δB,b(Q) implies δ-1

B,bδA,a = δ L,b

a .

Proof:

Let δA,a(P) = V [1] and δB,b(Q) = V [2]

From [1]: δA,a(P) = V and

AV = aAP

AV / AP = a

From [2]: δB,b(Q) = V

BV = bBQ

BV / BQ = b

By Lemma 1.2, if a = b, then the lines AB and PQ are parallel. However, it is given that the

lines AB and PQintersect at L.

Therefore, a ≠ b and hence b

a ≠ 1.

Therefore, by Theorem 3(Appendix C), δ-1

B,bδA,a = δ S,b

a for some point S on AB .

But δ-1

B,bδA,a (P) = δ-1

B,b(V) (from [1])

= Q. (from [2])

So the point S lies on the line PQ as well. It is given that the lines AB and PQ intersect at L.

Since the point S is on both AB and PQ it must be the point of intersection of these two lines.

Therefore, S = L.

Hence, we have δ-1

B,bδA,a = δ L,b

a .

CHAPTER 2

MENELAUS’ THEOREM

Theorem 2.1:

Suppose points D, E, F are respectively on lines BC , AC , AB and each is distinct

from the vertices of ∆ABC. Then points D, E, F are collinear iff

(AF / FB)(BD / DC)(CE / EA) = −1.

Proof:

Figure 2.1.1

Let f = FA / FB,

d = DB / DC,

and e = EC / EA.

Then, δF,f (B) = A, [1]

δD,d (C) = B, [2]

and δE,e (A) = C. [3]

Substituting [2] into [1] gives:

δF,f (δD,d (C)) = A

δF,f δD,d (C) = A [4]

Substituting [3] into [4] gives:

δF,f δD,d (δE,e (A)) = A

δF,f δD,d δE,e (A) = A.

So dilatation δ = δF,f δD,d δE,e fixes the point A.

Assume the points D, E, F are collinear.

Let l = DE .

Look at δF,f δD,d δE,e (l):

The point E is on the line l. This implies δE,e (l)= l.

Therefore, δF,f δD,d δE,e (l)= δF,f δD,d (l).

Similarly, the point D is on the line l. This implies δD,d (l)= l.

Therefore, δF,f δD,d δE,e (l) = δF,f δD,d (l)

= δF,f (l).

Chapter 2: Menelaus’ Theorem

Also, the point F is on the line l. This implies δE,e (l)= l.

Therefore, δF,f δD,d δE,e (l) = δF,f δD,d (l)

= δF,f (l)

=l.

Hence, dilatation δ = δF,f δD,d δE,e fixes the line l = DE .

We have a dilatation δ = δF,f δD,d δE,e which fixes the point A and the line DE . It is given

that the point A is not on the line DE . Therefore, δ fixes the line DE and a point A which

is off the fixed line. By part (a) of the lemma 1.1, dilatation δ must be the identity.

If it is the identity, then

δF,f δD,d δE,e = ι

δD,d δE,e = δ-1

F,f

= δF,f

1 .

By Theorem 2 (Appendix C) if de = 1, then δD,d δE,e is a translation.

But δD,d δE,e = δF,f

1 .

Therefore, the translation δD,d δE,e fixes the point F.

Therefore, it must be the identity translation.

Therefore, δF,f

1 = ι.

This implies f

1 = 1.

Therefore, f = 1.

This is a contradiction since δF,f (B) = A and A ≠ B (if f=1, then δF,f (B) = δF,1 (B) = B).

Hence, de ≠ 1.

We know δD,d δE,e = δF,f

1 .

If de ≠ 1, then by Theorem 3 (Appendix C):

δD,d δE,e = δF, de

implies

de = f

1

fde = 1.

We have,

(FA / FB)(DB / DC)(EC / EA) = fde = 1.

Chapter 2: Menelaus’ Theorem

By changing the directions of the line segments give us:

(−AF / FB)(−BD / DC)(−CE / EA) = (−f)(−d)(−e)

= −1.

Therefore, if the points D, E, F are collinear, then

(AF / FB)(BD / DC)(CE / EA) = −1.

Conversely, suppose that

(AF / FB)(BD / DC)(CE / EA) = −1.

(−f)(−d)(−e) = −1

−fde = −1

fde = 1.

Let δ = δF,f δD,d δE,e. Then δ is a dilatation. Since, δ fixes the point A, by Theorem 7 (Appendix

C), it is a dilation.

We have fde = 1.

Suppose fd = 1.

Then e = 1. This implies δE,e = ι.

Therefore, δ = δF,f δD,d ι.

But fd = 1, therefore, by Theorem 2 (Appendix C) δ is a translation.

But δ fixes the point A.

So, δ = ι.

Suppose fd ≠1.

Then, by Theorem 3 (Appendix C),

δF,fδD,d = δQ,fd for some point Q.

Therefore, δ = δF,f δD,d δE,e

= δQ,fd δE,e.

We know that fde = 1, therefore, by Theorem 2 (Appendix C), δ is a translation.

But δ also fixes the point A.

Therefore, δ must be the identity.

We have δ = δF,f δD,d δE,e = ι, whether or not fd = 1.

We also know that f ≠ 1 (as A ≠ B).

Therefore, by part (c) of the lemma 1.1, the points D, E, F must be collinear.

CHAPTER 3

CEVA’S THEOREM

Theorem 3.1:

Suppose points D, E, F are respectively on lines BC , AC , AB and each is distinct from

the vertices of ∆ABC. Then lines AD , BE , CF are concurrent or parallel iff

(AF / FB)(BD / DC)(CE / EA) = +1.

Proof:

The following three figures represent all cases.

Figure 3.1.1 Figure 3.1.2 Figure 3.1.3

Let f = FA / FB,

c = CA / CE,

d = DC / DB,

and a = AC / AE.

Then, δF,f (B) = A, [1]

δC,c (E) = A, [2]

δD,d (B) = C, [3]

and δA,a (E) = C. [4]

Let δ1 = δ-1

C,c δF,f and δ2 = δ-1

A,a δD,d.

Then δ1, δ2 are products of dilatations. Hence, by Theorem 5 (Appendix C) they are dilatations.

δ1(B) = δ-1

C,c δF,f (B) and

= δ-1

C,c (A) from [1]

= E

δ2(B) = δ-1

A,a δD,d (B)

= δ-1

A,a (C) from [3]

= E

Therefore, we have δ1(B) = δ2(B) = E. [5]

Suppose the lines AD , BE , CF are parallel.

By lemma 1.2, f = c if CF ║ BE and d = a if AD ║ BE .

This implies c

f = 1 and a

d = 1.

Chapter 3: Ceva’s Theorem

Therefore, c

f = a

d = 1 implies cd

fa = 1.

(FA / FB)(AC / AE)(CE / CA)(DB / DC) = 1

(-AF / FB)(-BD / DC)(AC / -EA)(CE / -AC) =1

implies (AF / FB)(BD / DC)(CE / EA) = +1.

Suppose the lines AD , BE , CF are concurrent.

Then line CF is not parallel to the line BE , hence f ≠ c, hence cf

≠ 1.

Therefore, by Theorem 3 (Appendix C)

δ1 = δ-1

C,c δF,f

= δP, c

f for some point P on the line CF .

And also line AD is not parallel to the line BE , hence d ≠ a, hence ad

≠ 1.

Therefore, by Theorem 3 (Appendix C)

δ2 = δ-1

A,a δD,d

= δQ, a

d for some point Q on the line AD .

From [5] we also have δ1 (B) = δP, c

f (B) = E and δ2 (B) = δQ, a

d (B) = E.

Therefore, the points P and Q both lie on the line BE as well.

Since the point P lies on BE and CF , the point of intersection of the lines must be P.

Similarly, since the point Q lies on BE and AD , the point of intersection of the lines must

be Q.

Since the lines AD , BE , CF are concurrent, the point P and Q must be the same.

Therefore, δ1 = δP, c

f and δ2 = δP, a

d .

δ1, δ2 are dilatations. Since δ1(P) = P and δ1(B) = E, and δ2(P) = P and δ2(B) = E,

δ1 = δ2 by Corollary 2 of Theorem 1 (Appendix C).


Since the centre of the dilations is the same, the ratios must be equal as well.

Therefore, we have c

f = a

d

implies (FA / FB)(CE / CA) = (DC / DB)(AE / AC)

(FA / FB) (DB / DC) (CE / CA) (AC / AE) = 1

(-AF / FB) (-BD / DC) (CE / CA) (-CA / -EA) = 1

implies (AF / FB)(BD / DC)(CE / EA) = +1.

Conversely, suppose

(AF / FB)(BD / DC)(CE / EA) = +1.

(AF / FB) (DB / DC) (AC / AE) (CE / AC) = 1

(−f)(−d

1 )(−a) (−c

1 ) = 1

( cf

)(d

a ) = 1,

so c

f = a

d .

If c

f = d

a = 1, then f =c and a = d, hence by lemma 1.2 the lines CF ║ BE and

AD ║ BE .

Therefore, the lines AD , BE , CF are parallel.

If cf ≠1 and

da ≠ 1, then we have f ≠ c and a ≠ d.

Therefore, δ1 = δ-1

C,c δF,f

= δQ, c

f (Theorem 3 [Appendix C]) for some point Q on the line CF ,

and δ2 = δ-1

A,a δD,d

= δR, a

d (Theorem 3 [Appendix C]) for some point R on the line AD .

From [5] we also have δ1 (B) = δQ, c

f (B) = E

and δ2 (B) = δR, a

d (B) = E.

Therefore, the points Q and R both lie on the line BE as well.

Since the point Q lies on BE and CF , the point of intersection of the lines must be Q.


Similarly, since the point R lies on BE and AD , the point of intersection of the lines must

be R.

But δQ, c

f (B) = δ1(B) = E

Therefore, QE = cf

QB

QE / QB = c

f

And δR, a

d (B) = δ2(B) = E

Therefore, RE = a

d RB

RE / RB = a

d

Since c

f = a

d , we have QE / QB = RE / RB.

This implies EQ / QB = ER / RB (=

− c

f

= −

ad ).

But points Q and R lie on BE .

By Theorem 4 (Appendix C), there is a unique point P on BE such that EP / BP =

− c

f ,

since −c

f ≠ 1.

Both Q and R are on BE and satisfy this equation. So Q = R by Theorem 4 (Appendix C),

since

−c

f ≠ − 1.

Hence, the lines AD , BE , CF intersect at the point Q and therefore are concurrent.

CHAPTER 4

DESARGUES’ THEOREM

Theorem 4.1:

Suppose points A, B, C, A΄, B΄, C΄, D, E, F, V are distinct and triangles ∆ABC and ∆ A΄B΄C΄

are such that lines BC and ''CB intersect at D, lines AC and ''CA intersect at E, lines

AB and ''BA intersect at F, and lines 'AA and 'BB intersect at V. Then lines 'AA , 'BB ,

'CC are concurrent iff points D, E, F are collinear. That is, triangles ∆ABC and ∆ A΄B΄C΄

are copolar iff they are coaxial.

Proof:

Figure 4.1.1

Assume ∆ABC, ∆A΄B΄C΄ are copolar. That is, the lines 'AA , 'BB , 'CC are concurrent

at point V.

Let a = (AV / AA΄),

b = (BV / BB΄),

and c = (CV / CC΄).

Then, δA,a (A΄) = V, [1]

δB,b (B΄) = V, [2]

and δC,c (C΄) = V. [3]

Suppose, a = b, then by lemma 1.2, the lines AB and ''BA must be parallel.

However, it is given that the lines AB and ''BA intersect at F.

Therefore, the lines AB and ''BA cannot be parallel. So, a ≠ b implies b

a ≠ 1.

Then δ-1

B,bδA,a = δP, b

a for some point P on the line AB (Theorem 3 [Appendix C]).

Chapter 4: Desargues’ Theorem

But δ-1

B,bδA,a (A΄) = δ-1

B,b (V) from [1]

= B from [2]

So the point P is on the line ''BA .

It is given that lines AB and ''BA intersect at the point F. Since P is on both AB and

''BA , it must be the point of intersection of these two lines.

Therefore, P = F.

Hence, δ-1

B,bδA,a = δF, b

a .

Suppose, b = c, then by lemma 1.2, the lines BC and ''CB must be parallel.

However, it is given that the lines BC and ''CB intersect at D.

Therefore, the lines BC and ''CB cannot be parallel. So, b ≠ c implies c

b ≠ 1.

Then δ-1

C,cδB,b = δQ, c

b for some point Q on the line BC (Theorem 3 [Appendix C]).

But δ-1

C,cδB,b (B΄) = δ-1

C,c (V) from [2]

= C΄ from [3]

So the point Q is on the line ''CB .

It is given that lines BC and ''CB intersect at the point D. Since Q is on both BC and

''CB , it must be the point of intersection of these two lines.

Therefore, Q = D.

Hence, δ-1

C,cδB,b = δD, c

b .

Suppose, c = a, then by lemma 1.2, the lines AC and ''CA must be parallel.

However, it is given that the lines AC and ''CA intersect at E.

Therefore, the lines AC and ''CA cannot be parallel. So, c ≠ a implies a

c ≠ 1.

Then δ-1

A,aδC,c = δR, a

c for some point R on the line AC (Theorem 3[Appendix C]).

But δ-1

A,aδC,c (C΄) = δ-1

A,a (V) from [3]

= A΄ from [1]

So the point R is on the line ''CA .

Chapter 4: Desargues’ Theorem

It is given that lines AC and ''CA intersect at the point E. Since R is on both AC and

''CA , it must be the point of intersection of these two lines.

Therefore, R = E.

Hence, δ-1

A,aδC,c = δE, a

c .

Therefore,

δE, a

c δD, c

b δF, b

a = δ-1

A,aδC,c δ-1

C,cδB,b δ-1

B,bδA,a

= δ-1

A,aδA,a

= ι.

So, δE, a

c δD, c

b δF, b

a = ι and a

c

,

cb

,

ba ≠ 1.

By part (c) of the lemma 1.1, the points D, E, F must be collinear. Hence, triangles ∆ABC and

∆ A΄B΄C΄ are coaxial.

Figure 4.1.2

Conversely, suppose the points D, E, F are collinear (see Figure 4.1.2).

Then triangles ∆AEA΄ and ∆ BDB΄ are copolar. Therefore, they must be coaxial by the first

part of the proof.

Hence, C΄, C, V are collinear. Hence, 'CC passes through V and triangles ∆ABC and

∆ A΄B΄C΄ are copolar.

CHAPTER 5

PAPPUS’ THEOREM

Theorem 5.1:

Suppose A, B, C, D, E, F are six points such that points A, C, E are collinear, points B, D, F

are collinear, lines AB and DE intersect at point L, lines BC and EF intersect at point M,

lines CD and FA intersect at point N, lines AB and CD intersect at point P, lines CD and

EF intersect at point Q, and lines AB and EF intersect at point R. Then points L, M, N are

collinear.

Proof:

Figure 5.1.1

Let e = EQ / ER,

d = DQ / DP,

c = CP / CQ,

b = BP / BR,

a = AR / AP,

and f = FR / FQ.

Then δE,e (R) = Q, [1]

δD,d (P) = Q, [2]

δC,c (Q) = P, [3]

δB,b (R) = P, [4]

δA,a (P) = R, [5]

and δF,f (Q) = R. [6]

Chapter 5: Pappus’ Theorem

From [2] and [1] we have δD,d (P) = δE,e (R)

δ-1

E,e δD,d (P) = R [7a]

It is given that the lines AB and DE intersect at point L. The line AB = PR since the

points P and R are on the line AB (see Figure 5.1.1).

Therefore, by lemma 1.3 δ-1

E,e δD,d = δL,e

d . [7b]

From [4] and [3] we have δB,b (R) = δC,c (Q)

δ-1

C,c δB,b (R) = Q [8a]

It is given that the lines BC and EF intersect at point M. The line EF = RQ since the

points R and Q are on the line EF (see Figure 5.1.1).


C,c δB,b = δM,c

b . [8b]

From [6] and [5] we have δF,f (Q) = δA,a (P)

δ-1

A,a δF,f (Q) = P [9a]

It is given that the lines CD and FA intersect at point N. The line CD = QP since the

points Q and P are on the line CD (see Figure 5.1.1).


A,a δF,f = δN,a

f . [9b]

From [7a] and [7b] we have δ-1

E,e δD,d (P) = δL,e

d (P) = R [10]


C,c δB,b (R) = δM,c

b (R) = Q [11]


A,a δF,f (Q) = δN,a

f (Q) = P [12]

By substituting [11] into [12] gives δN,a

f δM,c

b (R) = P [13]

By substituting [10] into [13] gives δN,a

f δM,c

b δL,e

d (P) = P.

By Theorem 5 (Appendix C) δN,a

f δM,c

b δL,e

d is a dilatation.

Since the dilatation δN,a

f δM,c

b δL,e

d fixes a point, namely point P, it is a dilation by

Theorem 7 (Appendix C). Since the point P is fixed by δN,a

f δM,c

b δL,e

d , the centre of

dilation must be P and the dilation ratio must be (fbd) / (ace).

So we have δN,a

f δM,c

b δL,e

d = δP,ace

fbd . [14]

Substituting [1] into [3] gives δC,c δE,e (R) = P [15]

Substituting [5] into [15] gives δC,c δE,e δA,a (P) = P

Chapter 5: Pappus’ Theorem

It is given that the points A, C, E are collinear.

Let l be the line AC .

Now look at δC,c δE,e δA,a (l):

Since A is on l we have δA,a (l)= l.

Therefore, δC,c δE,e δA,a (l) = δC,c δE,e (l)

= δC,c (l) [since E is on the line l]

=l [since C is on the line l].

So we have δC,c δE,e δA,a, which fixes point P and line AC . It is given that P is not on the line

AC . So we have δC,c δE,e δA,a, which fixes a line and a point which is off the fixed line (see

Figure 5.1.1). Therefore, by part (a) of lemma 1.1, it must be the identity.

Therefore, δC,c δE,e δA,a = ι implies (cea) = 1.

Substituting [4] into [2] gives δD,d δB,b (R) = Q [16]

Substituting [6] into [16] gives δD,d δB,b δF,f (Q) = Q

It is given that the points B, D, F are collinear.

Let l be the line BD .

Now look at δD,d δB,b δF,f (l):

Since F is on l we have δF,f (l)= l.

Therefore, δD,d δB,b δF,f (l) = δD,d δB,b (l)

= δD,d (l) [since B is on the line l]

=l [since D is on the line l].

So we have δD,d δB,b δF,f, which fixes point Q and line BD . It is given that Q is not on the line

BD . So we have δD,d δB,b δF,f, which fixes a line and a point which is off the fixed line (see

Figure 5.1.1). Therefore, by part (a) of lemma 1.1, it must be the identity.

Therefore, δD,d δB,b δF,f = ι implies (dbf) = 1.

Therefore, [14] becomes δN,a

f δM,c

b δL,e

d = δP,ace

fbd = ι.

We have δN,a

f δM,c

b δL,e

d = ι and also we have a

f ≠ 1 (if a

f = 1, then δN,a

f (Q) =

δN,1(Q) = Q. From [12] we have δN,a

f (Q) = P and also it is given that P ≠ Q). Therefore by

part (c) of lemma 1.1, the points L, M, N are collinear.

References

21

REFERENCES

Martin, G. E. (1997). Transformation Geometry – An Introduction to Symmetry. New York:

Springer-Verlag New York Inc.

Appendices

22

APPENDIX A

PROOF OF LEMMA 1.2 (CASE 2 AND CASE 3)

Case 2 (r > 1):

It is given that

δA,r (C) = E and

AE = r AC,

AE / AC = r

δB,s (D) = E

BE = r BD,

BE / BD = s

Suppose r = s:

Then we have

AE / AC = BE / BD


AE = x AC and also CE = AE – AC

AE / CE = x AC / (AE – AC)

= (x AC / AC) / ([AE / AC] – [AC / AC])

= x / (x – 1).

BE = x BD and also DE = BE – BD

BE / DE = x BD / (BE – BD)

= (x BD / BD) / ([BE / BD] – [BD / BD])

= x / (x – 1).

CE = AE – AC and

= xAC – AC

= (x – 1)AC

DE = BE – BD

= xBD – BD

= (x – 1)BD.

Appendices

23

Then

AB2 = AE

2 + BE

2 – 2AE BEcosθ

= x2AC

2 + x

2BD

2 – 2x

2AC BDcosθ

= x2 [AB

2 + BD

2 – 2AC BDcosθ]

and

CD2 = CE

2 + DE

2 – 2CE DEcosθ

= (x-1)2AC

2 + (x-1)

2BD

2 – 2(x-1)

2AC BDcosθ

= (x-1)2 [AB

2 + BD

2 – 2AC BDcosθ].

AB2 x

2 [AB

2 + BD

2 – 2AC BDcosθ]

=

CD2 (x-1)

2 [AB

2 + BD

2 – 2AC BDcosθ]

( AB/ CD) = x / (x – 1).


∆ ABE ~ ∆CDE.


m AEB = m CED

m BAC = m DCE

m ABD = m CDE





m AEB = m CED (common)

m BAC = m DCE (corresponding angles)

m ABD = m CDE (corresponding angles)


Appendices

24


Let a = AE / CE = BE / DE.

Then, we have

AE AE = = a

CE AE – AC

AE = a (AE – AC)

(a − 1) AE = a AC

AE a

= AC (a − 1)

and

BE BE

= = a

DE BE – BD

BE = a (BE – BD)

(a − 1) BE = a BD

BE a

=

BD (a − 1)

Therefore, AE / AC = a / (a – 1) = BE / BD.

This implies r = AE / AC = BE / BD = s.


Case 3 (r < 0):

Appendices

25

It is given that

δA,r (C) = E and

AE = r AC,

AE / AC = r

δB,s (D) = E

BE = r BD,

BE / BD = s

Suppose r = s:

Then we have

AE / AC = BE / BD


AE = x AC and also

= −x CA

=−x (CE − AE)

AE = x AE – x CE

x CE = x AE − AE

(CE / AE) = (x – 1)/ x

BE = x BD

= −x DB

=−x (DE − BE)

BE = x BE – x DE

x DE = x BE − BE

(DE / BE) = (x – 1)/ x

(CE / AE) = (x – 1)/ x and

CE = [(x – 1)/ x] AE

(DE / BE) = (x – 1)/ x

DE = [(x – 1)/ x] BE

Then

CD2 = CE

2 + DE

2 – 2CE DEcosθ

= [(x-1)/x]2AE

2+[(x-1)/x]

2BE

2–2[(x-1)

2AE BEcosθ

= [(x-1)/x]2 [AE

2 + BE

2 – 2AE BEcosθ].

and

AB2 = AE

2 + BE

2 – 2AE BEcosθ

CD

2 [(x-1)/x]

2 [AE

2 + BE

2 – 2AE BEcosθ]

=

AB2 AE

2 + BE

2 – 2AE BEcosθ

( CD/ AB) = (x – 1) / x.

Appendices

26


∆ ABE ~ ∆CDE.


m AEB = m CED

m EAB = m ACD

m EBD = m BDC





m AEB = m CED (common)

m EAB = m ACD (corresponding angles)

m EBD = m BDC (corresponding angles)



Let a = AE / CE = BE / DE.

Then, we have

AE AE

= = a

CE CA + AE

AE = a (CA + AE)

(1 − a) AE = a CA

AE a

=

CA (1 − a)

and

Appendices

27

BE BE

= = a DE DB + BE

BE = a (DB + BE)

(1 − a) BE = a DB

BE a

=

DB (1 − a)

Therefore,

AE / CA = a / (1 – a) = BE / DB

AE / −AC = BE / −BD.

This implies

−r = AE / −AC = BE / −BD = − s.

That is,

r = s.


Appendices

28

APPENDIX B

DEFINITIONS

Dilation:

A dilation about point P is either a stretch about P or a stretch followed by a halfturn about P

(Project 7 in this group).

Symbol: δP,r

δP,r is a dilation about point P with ratio r (r is called the dilation ratio). If r > 0 then δP,r is

the stretch about P of ratio r; if r < 0 δP,r which is defined by σPδP, r is the stretch about P of

ratio r followed by a halfturn about P.

Example:

δP,r(C)= C΄ (r > 0) and δP,-r(C)= C΄΄ (r < 0)

Note:

1. We suppose the lines in the plane are directed (in an arbitrary fashion) and AB

denotes the directed distance from A to B on the line AB . For any points A and B we

have AB = − BA.

2. P C΄ = rPC in the first case and P C΄΄ = rPC in the second case.

In general δP,r(A)= B iff PB = rPA.

3. If l is a line through P, then δP,r(l)= l, that is, any line through P is fixed by δP,r.

4. If δP,r(A)= B, then P lies on line AB .

Dilatation:

Dilatation is a collineation α such that l║α(l) for every line l (Project 7 in this group).

A fixed point:

We say transformation α fixes point P iff α(P) = P (Project 1 in this group).

Appendices

29

A fixed line:

We say transformation α fixes a line l iff α(l) = l (Project 1 in this group).

The identity:

The identity transformation ι is defined by ι(P)= P for every point P (Martin, 1997, p.7).

Copolar:

Figure 1

In Figure 1 ∆ABC and ∆ A΄B΄C΄ are said to be copolar as the lines 'AA , 'BB , 'CC joining

corresponding vertices of the triangles are concurrent at point V (Martin, 1997, p.150).

Coaxial:

Figure 2

In Figure 2 ∆ABC and ∆ A΄B΄C΄ are said to be coaxial as the points D, E, F of intersection of

corresponding extended sides of the triangles are collinear on line l (Martin, 1997, p.150).

Appendices

30

APPENDIX C

THEOREMS THAT WERE MADE USE OF

Theorem 1:

If AB ║ '' BA , then there is a unique dilatation δ such that δ(A) = A΄ and δ(B) = B΄

(Project 7 in this group).

Corollary:

1. A dilatation with two fixed points is the identity.

2. If two dilatations agree on two points, they must be the same.

Theorem 2:

Given δA,r

1 and δB,r, then for some point C

δB,r δA,r

1 = τA,C

[Theorem 13.14 (Martin, 1997, p.143)].

Theorem 3:

Given δA,r and δB,s, with rs ≠ 1, then for some point C

δB,s δA,r = δC,rs

[Theorem 13.14 (Martin, 1997, p.143)].

Theorem 4:

If y ≠ −1, then there exists a unique point P on AB different from B such that AP / PB = y

[Theorem 13.8 (Martin, 1997, p.140)].

Theorem 5:

Dilatations form a group. In particular, the product of two dilatations is a dilatation (Project

1 in this group).

Theorem 6:

A dilation is a dilatation (Project 7 in this group).

Theorem 7:

A dilatation with a fixed point is a dilation (Project 7 in this group).

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