elementary triangle geometry (mark dabbs)

7/29/2019 Elementary Triangle Geometry (Mark Dabbs)

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Elementary Triangle GeometryMark Dabbs

The Mathematical Association ConferenceUniversity of York, U.K

Spring 2004

Version 1.1 April 2004(www.mfdabbs.com)


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Contents

Motivating Problem. 5

1: Basic Trigonometrical Formulae. 11

2: Further Trigonometrical Formulae. 13

3: Ratio Theorems. 15Theorem (3.1).

Theorem (3.2).

4: Basic Triangle Formulae. 17Cosine Rule.Sine Rule and Cicumcircle.

Tangent Rule.

5: Other Triangle Formulae. 21Area Formulae in terms ofa, band c.

Sin A, sin B and sin C in terms ofa, band c.

Cos A, cos B and cos C in terms ofa, band c.

Tan A, tan B and tan C in terms ofa, band c.

6: Associated Circles. 25Incircle.

Excircles.Herons Area Formula.

7: Further Triangle Formulae. 33

( ) ( ) ( )1 12 2 2an , tan and tan in terms of , and .A B C a b c ( ) ( ) ( )1 12 2 2os , cos and cos in terms of , and .A B C a b c ( ) ( ) ( )1 12 2 2in , sin and sin in terms of , and .A B C a b c

8: Further Triangle Relationships. 37

elationship between and .Rr elationships between , , and .

A B Cr r r R

he distances A , and C .I BI I he distances A , and C .

A B CI BI I

he distances , and .A B C

II II II

9: Further Triangle Centres. 47The Orthocentre of any Triangle ABC.

The Pedal Triangle of any Triangle ABC.

The Circumcircle and the Pedal Triangle.The Excentric Triangle.


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10: Special Cevian Lengths. 55The Centroid and Medians of any Triangle.

Cevians Bisecting Angles Internally.

Cevians Bisecting Angles Externally.

Apollonius Theorem.A Generalisation of Apollonius Theorem Stewarts Theorem

11: Problems. 63

Appendix: Concurrences of Straight Lines in a Triangle. 67Circumcentre.

Incentre.

Centroid.

Orthocentre.

Bibliography. 73


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Motivating Problem

a ch

x b - x

PC

B

A

The motivation for this work came from an open question to a class to find the

area of a triangle whose base is known but whose perpendicular height is not

known.A typical diagram is shown in Figure MP.1

Figure MP.1

The area of the triangle, , is seen to be

( ) ( )12

12

base height

,bh

=

=(MP.1)

where and h .b AC= BP=

After some discussion, two methods were proposed

o Method 1: Using Trigonometry

o Method 2: Using Pythagoras

Method 1 is perhaps the more familiar and progresses thus:

In triangle BPCwe have:

sin

sin

PBC

BC

h

C a

=

=

Therefore, sinh a C= (MP.2)

From (MP.1) and (MP.2) we have the general area formula

12 sinab C = (MP.3)


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Method 2 was somewhat more involved and led to quite a voyage of discovery!

Note the following Pythagorean relations within the two triangles CBPand ABP.

2 2

a x h

2

= + (MP.4)and ( )

22c b x h2= + . (MP.5)

Eliminating hfrom (MP.4) and (MP.5) gives:

( )22 2 2a x c b x =

That is: .( )22 2 2a c x b x =

Thus, by the difference of two squares formula we have

( ) ( )

( )(

( ) ( )

2 2

2 2

2 2 2

a c x b x x b x

a c x b x x b x

a c x b b

= +

= + +

=

)

Hence,2 2

2

a b cx

b

2+ = . (MP.6)

Substituting (MP.6) into (MP.4) gives22 2 2

2 2

2

a b ch a

b

+ =

.

That is

( )22 2 2 2 2

2

2

4

4

a b a b ch

b

+ = (MP.7)

Once again, by the difference of two squares formula we have the alternative form

of(MP.7):

( )( )2 2 2 2 2 22

2

2 2

4

ab a b c ab a b ch

b

+ + + = ,

( ) ( )2 2 2 2 2 22

2 2

4

ab a b c ab a b c

b

+ + + = ,

( ) ( )2 2 2 2 2 2

22 2

4c a ab b a ab b c

b + + + = ,


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That is:

( )( )2 2 2 2 2 22

2

2 2

4

c a ab b a ab b ch

b

+ + + = ,

which, on factorising gives:

( )( ) ( )( )2 22 22

24

c a b a b ch

b

+ = .

Therefore, on using the difference of two squares formula again we have:

( )( ) ( ) ( )22

4

c a b c a b a b c a b ch

b

+ + + +=

( )( )( ) ( )224

c a b c a b a b c a b ch

b

+ + + + += . (MP.8)

Now its time to ask which of the four factors in the numerator looks the nicest

and hope that the answer to come back is the fourth or last one of ( ) !a b c+ +

Having established this, the suggestion is then made that it is a pity that the other

three factors do not have this same elegant symmetry and once agreed that we

ought to insistthat such symmetry exist in these other three factors.

It is eventually determined that a suitable trick is to rewrite them in the

following manner:

( ) ( )

( ) (

( ) (

2

2

2

c a b a b c a

c a b a b c b

a b c a b c c

+ + +

+ + +

+ + +

)

)

)

(MP.9)

Realising that (a b c+ + is just the perimeter of the original triangle ABC, sayp

gives (MP.8) as:

( )( )( )( )22

2 2 2

4

p a p b p c ph

b

= . (MP.10)

However, if we then let the new variable, s, be defined as the semi-perimeterthen

(MP.10) is re-written

( ) ( ) ( ) ( )22

2 2 2 2 2 2 2

4

s a s b s c sh

b

= .


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This is then easily factorised to give:

( )( )( )( )22

16

4

s a s b s c sh

b

=

or( )( )( )2

2

4s s a s b s ch

b

= .

Hence,( )( ) ( )2 s s a s b s c

hb

= (MP.11)

Finally then, substituting (MP.11) back into (MP.1) gives

( )( ) ( )12 2 s s a s b s cbb

=

or

(MP.12)( )( )( )s s a s b s c =

Which is the familiar result of Heron of Alexandria (First Century A.D)


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What a marvel that so simple a figure as the triangle is so

inexhaustible in its properties!

(A. L. Crelle, 1821)


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1: Basic Trigonometrical Formulae

sintan

cos

= (1.1)

2 2sin cos 1 + = (1.2)

0 30 45 60 90

sin 01

2

2

2

3

2 1

cos 13

2

2

2

1

2

0

(1.3)

tan 01

3 1 3

1cosec

sin1

seccos

1cot

tan

=

=

=

(1.4)

2tan 1 sec2 + = (1.5)

21 cot cosec2 + = (1.6)


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2: Further Trigonometrical Formulae

(2.1)( )sin sin cos cos sinA B A B A B+ = +

(2.2)( )sin sin cos cos sinA B A B A B =

(2.3)( )cos cos cos sin sinA B A B A B+ =

(2.4)( )cos cos cos sin sinA B A B A B = +

( )tan tan

tan1 tan tan

AA B

A B

B++ =

(2.5)

( )tan tan

tan1 tan tan

AA B

A B

B =

+(2.6)

sin sin 2sin cos2 2

P Q P QP Q

+ + =

(2.7)

sin sin 2cos sin2 2

P Q P QP Q

+ =

(2.8)

cos cos 2cos cos2 2

P Q P QP Q

+ + =

(2.9)

cos cos 2sin sin2 2

P Q P QP Q

+ =

(2.10)

( )sin 2 2sin cosA A A= (2.11)

(2.12)( ) 2 2 2cos 2 cos sin 2cos 1 1 2sinA A A A= = = 2A

( )( )2 12sin 1 cos 2A = A (2.13)

( )( )2 12cos 1 cos 2A = + A (2.14)

Notice further, if we define 12tant = then it can be shown that

2

2 2

2 1sin , cos and tan .

1 1

t t

t t

= = =

+ +2

2

1

t

t

(2.15)


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3: Ratio Theorems

Theorem (3.1)If we have that p a

q b= then p q a , for any numbersb

mp nq ma nb + +=

+ +, , and .m n

Proof:

Letp a

tq b

=

, .p qt a bt= =

Therefore, ( )( )qt qp q t

mp nq m qt nq mt n ++ += =+ + +

and

( )

( )

bt qa b t

ma nb m bt nq mt n

++ += =

+ + +

Theorem (3.2)

If we have that then , for any numbersp aq b=

p a p aq b q b

+= =+

and .

Proof:

Letp a

tq b

=

, .p qt a bt= =

Therefore,( ) ( ) ( )

( )

qt bt t q bp a ptq b q b q b q

a

b

+ ++

= = = + + +


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c ah

b

x y

A

B

C

4: Basic Triangle Formulae

Cosine Rule

Figure 4.1

Notice that: cos , cosx y

A Cc a

= =

Therefore, cos , cosx c A y a C= =

Hence cos cosb x y c A a C = + +

Using symmetry we interchange the variables to yield the complete set of results

thus:cos cos

cos cos

cos cos

a b C c B

b c A a C

c a B b A

= +

= +

= +

(4.1)

The formulae of(4.1) are known as the Projection Formulae.If we now multiply the equations of(4.1) by a, band c, respectively, we have:

(4.2)2 cos cosa ab C ac= + B

A

A

(4.3)2 cos cosb bc A ab C = +

(4.4)2 cos cosc ac B bc= +

Now construct (4.3) + (4.4) - (4.2) to give:

2 2 2 2 cosb c a bc+ =

Therefore,

(4.5)2 2 2 2 cosa b c bc= + A

a

bc

Equation (4.5) is known as the Cosine Rule for triangles.Symmetry yields the other forms:

2 2 2 2 2 22 cos and 2 cosb c a ca B c a b ab= + = + C


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Sine RuleAlso from Figure 4.1 we have the further set of relations:

sin , sinh h

A C

c a

= =

Therefore, sin or sinh c A h a C = =

Hence, sin sinc A a C h=

Therefore,sin sinA C

a c=

However, the initial orientation of the triangle ABC was arbitrary

(4.6)sin sin sinA B C

a b c= =

Equation (4.6) is known as the Sine Rule for triangles.The Sine Rule can be extended by considering a circle through the apexes of the

triangle ABC (known as the Circumcircleof the triangle ABC)

P

P

OO

B

A C

A

B

C

Figure 4.2 Figure 4.3

In both Figure 4.2 and 4.3 the red lines AP and PC have been added to the

original Circumcircle problem. In both cases the line segment AP is draw so as to

pass through the centre of the Circumcircle and is therefore a diameter.

ACP is a Right-Angle in both figures (Angle in a Semi-Circle is a right-

angle).

Further, APC = ABC since angles subtended by a single chord in the same

segment of a circle are equal (Euclid Book III Prop. 21).

Therefore, from Figure 4.2 we have: ( ) ( )sin sin2

AC bAPC BAP R

=

where Ris the radius of the Circumcircle of triangle ABC.


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From Figure 4.3 180APC B= (Cyclic Quadrilateral)

Therefore, ( ) ( )sin sin 180 sinAPC B B=

Hence, as for Figure 4.2 we have

sin 1sin or

2 2

b BB

R b R= =

Therefore, from (4.6) we have:

sin sin sin 1

2

A B C

a b c(4.7)

R= = =

Tangent RuleFrom (4.7) we have that

sin

sin

b

c C=

B

Using Theorem (3.1) with and1 1m n = = this relation can be written

( )

( )

( ) ( )

( ) ( )

sin1 sin

1 1 in

1

s

1B

C

+ +

+

=

+

( )

( )

1

1 1 sin

b c

b c B

C

That issin sin

sin sin

b c B C

b c B C

=

+ +

Using (2.8) and (2.7) this can be rewritten as

2cos sin2 2

,

2sin cos

2 2

sin12

,

cos sin2 2

cos2

tan

2 .tan

2

B C B C

b c

B C B Cb c

B C

b c

B C B Cb c

B C

B C

b c

B Cb c

+ =

+ +

=

++

+

= ++


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That is ( ) ( )1 12 2tan tanb c

B C Bb c

C

=+

+ (4.8)

However, 180A B C+ + = ( )12 9012B C+ =

A (4.9)

From (4.9) we see that

( ) ( )

( )

1 12 2

12

12

12

12

12

1 12 2

tan tan 90

tan(90 ) tan

1 tan(90 ) tan

tan1

tan(90 )1

tantan(90 )

1 00 tan

tan cot

B C A

A

A

A

A

A

B C A

+ =

=

+

=+

=+

+ =

Hence (4.8) can be written in its alternative form

(4.10)( )12tan cotb c

B Cb c

=

+12 A

Equation (4.10) is known as the Tangent Rule for triangles.Symmetry yields the other forms:

( )

( )

1 12 2

1 12 2

tan cot

tan cot

c aC A B

c a

a bA B C

a b

=

+

=

+

a

bc


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5: Other Triangle Formulae

c ah

b

A

B

C

Area Formulae in terms of a, band c.

Figure 5.1

The area of triangle ABCis found from

( )

12

12

, where sin ,

sin

hbh C

ab a C

= =

=

Hence, 12 sinab C = (5.1)a

a

bc

By symmetry, 1 1 12 2 2sin sin sinab C bc A ca B = = =

From (4.7) we have that sin2

cC

R=

Therefore, (5.1) becomes 122

cab

R

=

That is:4

abc

R = (5.2)

bc

or4

abcR =

.

Also, from (4.7) we have that sinsin

a BbA=

Therefore, (5.1) now becomes:

12

sinsin

sin

a Ba C

A

=

That is: 212sin sin

sin

B Ca

A =

Hence,

2 2 21 1 12 2 2

sin sin sin sin sin sin

sin sin sin

B C C A Aa b c

A B = = =

B

C(5.3)


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SinA in terms of a, band c, etc.From (1.2)

( ) ( )( ) (

2 2

2 2

sin 1 cos

1 cos

1 cos 1 cos

A A

A

A A

=

= = + )

Therefore, from (4.5) this becomes

( )

2 2 2 2 2 22

2 2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2 2

22

sin 1 12 2

2 2

2 2

2 2

2 2

2 2

2 2

2

b c a b c aA

bc bc

bc b c a bc b c a

bc bc

bc b c a bc b c a

bc bc

a b bc c b bc c a

bc bc

a b c

bc

+ + = +

+ + + =

+ + + =

+ + + =

=

( )

( )( ) ( )( )

( )( )( )( )

( )( ) ( )(

2 2

2 22 2

2 2

2 2

2 2

2

1

4

1

41

2 2 24

b c a

bc

a b c b c ab c

a b c a b c b c a b c ab c

a b c b a b c c b c a a b c ab c

+

= +

= + + + + +

= + + + + + + + )+

Now let (12s a b= + + )c , the Semi-perimeter, then we have

( )( )( )( )

( )( )( )( )

2

2 2

2 2

1sin 2 2 2 2 2 2 2

4

1 164

A s b s c s ab c

s b s c s a sb c

=

=

s

Hence

( )( )(22 2

4sin A s s a s b s

b c= )c

Therefore,

( )( )( )2

sin A s s a s b s cbc

=

( )( )( )

2

sinB s s a s b s cca=

( )( )( )2

sinC s s a s b s cab

=

(5.4)

a

bc


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Notice further that these three identities from (5.4) could be written

( )( )( )

( )( )( )

( )( )( )

2sin 2

2sin 2

2sin 2

A s s a s b s c

B s s a s b s c Ka

aa

C s s a s b s cab

bb

cc

=

=

=

(5.5)

Kbca

cb

cK

From which, the Sine Rule can be deduced, sincesin sin sin 1

22

A B CK

a b c

= = R

Moreover, from (5.1) and (5.4) we have that

( )( )( )1 12 22

sinab C ab s s a s b s cab

=

Hence

(5.6)( )( )( )s s a s b s c =

This is the triangle area formula met previously in (MP.7): Herons Formula.

We can now use the notation of(5.6) or more simply the form of(5.1) to write:

SinA, sinB, sin C in terms of a, band c.

a

bc

(5.7)

2 2n , sin , sinA B C

bc ca ab

= = =si

a

bc

2

CosA, cosB, cos C in terms of a, band c.From (4.5) we simply rearrange to yield

(5.8)2 2 2 2 2 2 2 2 2

, cos , cos2 2

b c a c a b a b cA B C

bc ca ab

+ + + = = =cos

2

a

bc

TanA, tanB, tan C in terms of a, band c.From (1.1), (5.5) and (5.8) we have

(5.9)2 2 2 2 2 2 2 2 2

4 4tan , tan , tanA B C

b c a c a b a b c

= = =

+ + +

4


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6: Associated Circles

EF

D

I

BC

A

IncircleLet I be the Incentre of thetriangle ABC, obtained by

bisecting the interior angles of

the triangle ABC. Then

,ID IE IF r= =

( , and areID IE IF

where ris the

radius of the incircle.

.the 's

from to the respective sides)I Figure 6.1

We have that( )

( )

1 1 12 2 2

12

Area of triangle Areas of trianglesABC BIC CIA AIB

ar br cr

r a b c

= +

= + +

= + +

+

Hence fors, the semi perimeter

rs = (6.1)

or r

s

= (6.2)

Now, let

AE AF x

BD BF y

CD CE z

=

= = =

Tangents to a circle from

a single point have equal

lengths

( )perimeterx y y z z x p+ + + + + =

Hence

( )semi perimeterx y z s+ + =

But BD DC y z a+ = +

a s+ =

Hence x s a= .

Therefore, we can show:s a

y s bz s c

=

= =

(6.3)

a

bc


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From triangle ABC we see that

( )12ntaIF r

AFA s a

= =

Similarly( )

( )

( )

12

12

12

tan

n

tan

rA

s a

rB

s b

rC

s c

=

=

=

ta (6.4)

Moreover, from triangles AIE, BIF and CID we have

( )

(

( )

12

12

12

cot

cot

cot

AE AF s a r A

BD BF s b r B

CD CE s c r C

= =

= =

= =

(6.5))

Figure 6.2

s - bs - c

s - c

s - as - a

s - b

E

F

D

I

B

C

A

a

bc

a

bc

Therefore, as thena BD DC = +

( ) ( )( ) ( )( )

( )

( )

( )

( )

( ) ( ) ( ) ( )

( ) ( )

( )

( ) ( )

1 1

2 2

1 12 2

1 12 2

1 12 2

1 1 1 12 2 2 2

1 12 2

1 12 2

1 12 2

cot cotcot cot

cos cos

sin sin

cos sin sin cos

sin sin

sin

sin sin

a r B r C r B C

B Cr

B C

B C Br

B C

B Cr

B C

= +

= +

= +

+=

+=

C

But and therefore,180A B C+ + = 1 1 12 2 290B C A+ = , giving

( )( ) ( )

12

1 12 2

sin 90

sin sin

Aa r

B C

=

That is

( )

( ) ( )

12

1 12 2

cos

sin sin

Aa r

B C= .

Hence( ) ( )

( )

1 12 2

12

sin sin

cos

B Cr a

A= (6.6)


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27

Together, we have

( ) ( )

( )

( ) ( )( )

( ) ( )

( )

1 12 2

12

1 12 2

12

1 12 2

12

sin sin

cos

sin sin

cos

sin sin

cos

B Cr a

A

C Ar b

B

A Br c

C

=

=

=

(6.7)

From (4.7) we have

sin 1

2

A

a R= , etc, where Ris the Circumcircle.

or

a R2 sin A= , etc

Thus (6.6) becomes

( ) ( )

( )

1 12 2

12

sin sin2 sin

cos

B Cr R A

A=

Using (2.11) gives

( ) ( )( ) ( )

( )

1 12 21 1

2 2 12

sin sin2 2sin cos

cos

B Cr R A A

A=

Hence

(6.8)( ) ( ) ( )1 1 1

2 2 24 sin sin sinr R A B C =

a

bc


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Excirclesxcircles

A'D

I

D'

E'

F'

IA

B C

A

( )ABI C

Let AI be the centre of a circle, opposite

angle A, of radius obtained by bisecting

the angles Band C externally and Ainternally.

ArAr

Let AI be the centre of a circle, opposite

angle A, of radius obtained by bisecting

the angles Band C externally and Ainternally.

We have that:We have that:

Area of triangle

Areas of triangles A A

ABC

I B CI A

=

+

( )

Area of triangle

Areas of triangles A A A

ABC

I B CI A BI C

=

+

That isThat is

( )

( )

1 1 1

2 2 212

12

2

A A

A

A

cr br ar

r c b a

r a b c a

= +

= +

= + +

A

Hence fors, the semi perimeterHence fors, the semi perimeter

(6.9)( )Ar s a = (6.9)( )Ar s a =

A

B

rs a

rs b

=

=

Similarly, we haveSimilarly, we have

Crs c

=

(6.10)

A

B

C

rs a

rs b

rs c

=

=

=

(6.10)

Notice further that are collinear as both lie on the bisector

line of angle A

Notice further that are collinear as both lie on the bisector

line of angle A

, and AA I I, and AA I I and AAI AIand AAI AI

Now letNow let' '

' '

' '

A

A

A

AE AF x

BD BF y

CD CE z

=

= = =



lengths



lengths

' '

' '

' '

A

A

A

AE AF x

BD BF y

CD CE z

=

= = =

ThenThen ' ' 'AF AE AB BF AC CE+ = + + +' ' 'AF AE AB BF AC CE+ = + + + '

A

( )

2 A A

A A

x c y b z

b c y z

= + + +

= + + +

( )

2 A A

A A

x c y b z

b c y z

= + + +

= + + +

but .but .' 'A Ay z BD D C BC a+ = + =' 'A Ay z BD D C BC a+ = + =

a

bc

28

a

bc

'

A

Figure 6.3

A'D

I

D'

E'

F'

IA

B C

A

( )

( )

1 1 1

2 2 212

12

2

A A A

A

A

cr br ar

r c b a

r a b c a

= +

= +

= + +


29/77

29

Therefore:

( )12

2 A

A

a b c

a b c

= + +

= + +

Hence

( )semi perimeter (6.11)Ax s=

A'D

D'

E'

F'

IA

B C

A

In a similar manner

andB Cs x = x =

a

bc

s

The three distances , andA B Cx s x s x s= = =

are displayed below

Figure 6.4

Figure 6.5

IB

B C

A

Figure 6.6

IC

B C

A


30/77

30

Notice also from triangle and from (6.11) that'AAI F

( )12'

tan'

A AI F rAAF s

= =

( )12tanAr s A= .

Similarly

( )

(

( )

12

12

12

tan

tan

tan

A

B

C

r s A

r s B

r s C

=

=

=

(6.12))

a

bc

Therefore, we can further show that

( )( )

( )

12

12

12

' ' cotcot ,

cot .

A

B

C

AF s r As r B

s r C

= ==

=

(6.13),AE

B

a

bc

Moreover, since ' ' 'F BD AF AB s c= = =

and ' ' 'CE CD AE AC s b= = =

then

( ) (

' 'DD CD CD

s b s c

b c

)

=

=

=

Where D is the point of tangency of the incirclewith side a, is the point of

tangency of the excirclewith side a, and where the symbol is the positive

difference between its two arguments. Namely:

'D

'DD b c b c= (6.14)

Further, as 'BD CD s= = c it follows that is the midpoint of where

is also the midpoint ofBC.

'A 'DD 'A


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31

Herons Area Formula

IA

D

ID'

AB

C

Figure 6.7

As and ABI B

90AIBI =

I bisect the angle B both internally and externally, it follows that

. Moreover, since and 'AIDB I D B

'A

are also right angles then this

implies that triangles andBID BI D are similar.

Namely:Triangles '

A

BID I BD

Thus' 'A A

BI ID BD

I B BD I D= =

that is:A A

BI r s b

I B s c r

= =

Therefore: ( )( )Ar r s b s c = (6.15)

On using (6.2) and (6.10) this becomes

( )(s b s cs s a

) =

Hence ( )( )( )s s a s b s c = (6.16)

Which is the form of the area of triangle ABCas met previously in (MP.7) and

(5.6)


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32


33/77

33

7: Further Triangle Formulae

Figure 7.1

Y

D

I

IA

XA

C

B

( ) ( ) ( )1 12 2 2an , tan and tan in terms of , and .B C a b c

From Figure 7.1 we have ( )12tanID r

AAD s a

=

On using (6.15) this becomes

( )

( )( )

( )(

( )

)12tan

A

A

s b s c

r s b s cA

s a r s a

= =

On using (6.9) we have

( )( )( )

( )( )

( )( )12tan

s b s c s b s cA

s as a

= =

Hence: ( )( )( )

( )( )( )

( )(

( )

)12tan

s b s c s b s cA

s s as s a s b s c

= =

Similarly,

(7.1)

( )( )( )

( )

( )( )( )

( )

( )( )( )

( )

12

12

12

tan

tan

tan

s b s cA

s s a

s c s aB

s s b

s a s bC

s s c

=

=

=

a

bc


34/77

34

( ) ( ) ( )1 12 2 2os , cos and cos in terms of , and .B C a b c From Figure 7.1 we have ( )12 180AI CY C=

( )

1

212

90 C

A B

=

= +

Also, using the construction properties of the Incircle and Excircle

12A

I AC A=

Hence

( )( )

( )( )( )

( )

12

1 12 2

12

12

180 180

180 180

180 180

A A

A

AI C A I CY

A A

B

AI C B DBI

= +

= + +

=

=

B

Therefore, triangles are similar, that isand AAIB ACI AAIB ACI .

A A

A A

AI IB AB

AC CI AI

AI IB c

b CI AI

= =

= =

Hence AAI AI bc = (7.2)

Also from Figure 7.1 ( )12cosA

AXA

AI= (7.3)

and ( )12cosAD

AAI

= (7.4)

The product of(7.3) and (7.4) gives

( ) ( )1 12 2cos cosA

AX ADA A

AI AI =

That is

( )2 1

2cos A

AX AD

A AI AI

=

On using (6.3), (6.11) and (7.2) this becomes

( )( )2 1

2coss s a

Abc

=

Hence

(7.5)

( )( )

( )

( )

( )( )

12

1

2

12

cos

cos

cos

s s aA

bc

s s b

B ca

s s cC

ab

=

=

=

a

bc


35/77

35

( ) ( ) ( )1 12 2 2in , sin and sin in terms of , and .B C a b c Using (1.1), (7.1) and (7.5) we have that

( ) ( ) ( )1 12 2sin tan cosA A=12

A

Hence

( )( )( )

( )

( )12sin

s b s c s s aA

s s a bc

=

Giving

(7.6)

( )( )( )

( )( )( )

( ) ( )( )

12

12

12

sin

sin

sin

s b s cA

bc

s c s aB

ca

s a s bC

ab

=

=

=

a

bc

Notice that in each of the above three subsections the negative root is rejected if the

angles,A, B and Care those of a triangle.


36/77

36


37/77

37

8: Further Triangle Relationships

Relationship between r and RNote that

( )

( )

1 12 2

12

12

12

180

180

180 180

90

BIC B C

B C

A

A

=

= +

=

= +

r

D

I

A

B C

Figure 8.1

and that ( )12sinID r

BIB IB

=

Hence, ( )

1

2sinr IB B=

(8.1)Using the Sine Rule of (4.6) we have that

( ) ( ) ( )11 122 2sin 90sin sinAC BIB BC I

+= =

C,

and from the first two of these fractions we have

( ) ( )1 12 2sin cosC A

IB BC= .

Therefore: ( )( )

1

212

sincos

BC CIBA=

,

that is( )

( )

12

12

sin

cos

a CIB

A= . (8.2)

Substituting (8.2) into (8.1) gives

( )

( )( )

( ) ( )( )

12 1

212

1 1

2 212

sinsin

cos

sin sincos

a Cr B

A

a C BA

=

=

Using (4.7) acan be written as: a R2 sin ,A= giving( ) ( )

( )

1 12 2

12

2 sin sin sin

cos

R A Br

A=

C.

Using (2.11) gives

( ) ( ) ( ) ( )

( )

1 1 1 12 2 2 2

12

2 2sin cos sin sin

cos

R A A Br

A

=

C.

Hence: 1 1 1r R (8.3)( ) ( ) ( )2 2 24 sin sin sinA B C=

as met previously in (6.8).


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38

Relationships between , , and .A B Cr r r R

rA

D'IA

A B

C

From Figure 8.2 we see that

( )'

n 'A A

A

A Asi

I D r

I BD I B I B=

that is ( )12sin 90A Ar I B B=

(8.4) ( 12cosA Ar I B B= )

Figure 8.2

Using the Sine Rule of (4.6) we have that

( ) ( ) ( )1 12 2sin 90 sin 90 sin 90

A A

C A

I B BC I C

= =

12 B

and from the first two of these fractions we have

( ) ( )1 12 2cos cos

A

C A

I B BC=

Therefore,( )

( )

12

12

cos

cosA

BC CI B

A= .

that is,

( )

( )

12

12

cos

cosA

a C

I B A=

(8.5)

Substituting (8.5) into (8.4) gives

( )

( )( )

( ) ( )

( )

12 1

212

1 12 2

12

coscos

cos

cos cos

cos

A

a Cr B

A

a C B

A

=

=

Using (4.7) acan be written as: 2 sina R A= ,

giving( ) ( )( )

1 12 2

12

2 sin cos cos

cosA

R A Cr

A=

B.

Using (2.11) gives

( ) ( ) ( ) ( )

( )

1 1 1 12 2 2 2

12

2 2sin cos cos cos

cosA

R A A Cr

A

=

B.

Hence: ,a ( ) ( ) ( )1 1 12 2 24 sin cos cosAr R A B C =

( ) ( ) ( )1 1 12 2 24 cos sin cosBr R A B C =

( ) ( ) ( )1 1 12 2 24 cos cos sinC A B C=

and similarly , (8.6)

r R .

bc


39/77

39

XD

I

IA

A B

C

The distances A , and C .BI I

From Figure 8.3 notice that

( )12nsiIDAIA

=

Figure 8.3Therefore,

( )

(

( )

12

12

12

cosec

cosec

cosec

AI r A

BI r B

CI r C

=

=

=

(8.7))

a

bc

XD

I

IA

A B

If we just consider the triangle for the present and determine some of its

angles from Figure 8.4

we find that:

AAI X

Clearly ( )12180AI B A B= +

and so on using the Sine Rule of (4.6)

we have from triangle AIB

( ) ( ) ( )1 12 2sinsin sinAIBB AAI AB BI

= =

Figure 8.4

Considering just the first two fractions here gives

( )

( )

12sin

sin

AB BAI

AIB=

Therefore

( )( )( )

12

12

sin

sin 180

c BAI

A B=

+

That is on using (2.2):

( )( )( )

( )

( )( )

( )

12

12

12

12

12

12

sin

sin

sin

sin 90

sin

cos

c BAI

A B

c B

C

c B

C

=

+

=

=

But on using (4.7) c can be written as: 2 sin ,c R C= giving


40/77

40

( )

( )

12

12

2 sin sin

cos

R C BAI

C=

Using (2.11) gives( ) ( ) ( )

( )

1 1 12 2 2

12

2 2sin cos sin

cos

R C CAI

C

=

B

Hence

(8.8)

( ) ( )

( ) ( )

( ) ( )

1 12 2

1 12 2

1 12 2

4 sin sin

4 sin sin

4 sin sin

AI R B C

BI R C A

CI R A B

=

=

=

a

bc

which give an alternative form of(8.7). Further, from (8.7), as it is somewhat

easier we may proceed ( )( )

12 1

2

cosecsin

rAI r A

A= =

Thus, from (7.6) we have

( )( ) ( )( )

( )

( ) ( )( )

r r bAI

s b s c s b s c

bc

r s s a bc

s s a s b s c

= =c

( ) ( ).

r s s a bc s s a bcAI

s

= Using (5.6) and (6.1)

( ) ( )s a bc s abc

s s

= =

( ) ( )a s b c s bca

s s

= =

( ) ( )ab s c s cab

s s

= =

AI

BI

a

bc

similarly (8.9)

CI

Note that the relations of(8.9) can be written

( ) ( )

( ) ( )

( ) ( )

,

, where

.

s a s aa

a

b

b

c

c

bcAI K

s a

s b s ba c abcK K

s b

s c s cabCI K

s c

=

=

=

BI (8.10)s

=

(8.11)( ) ( ) ( )

: : : :s a s b s c

AI BI CIa b c

=

Hence,


41/77

41

XD

I

IA

A B

C

The distances A , and C .A B C

BI I

Figure 8.5From Figure 8.5 notice that

( )12sinA

A

I DA

I A

=

Therefore,

( )

( )

( )

12

12

12

cosec

cosec

cosec

A A

B B

C C

AI r A

BI r B

CI r C

=

=

=

(8.12)

a

bc

D

I

IA

B

C

D

I

IA

A B

Notice also from Figure 8.6 that

180

90

+ + + =+ =

Hence

90AIBI =

Also, since

( )1 12 2and 180 90 12B B B = = =

then ( )1290AI I B A B= +

Figure 8.6

Figure 8.7


42/77

42

On using the Sine Rule of (4.6) we have from triangle AABI

( ) ( ) ( )1 12 2sin 90 sin sinA

A A

B AI B A

AI AB BI

+= =

Considering just the first two fractions here gives

( )( )

12sin 90

sinA

A

AB BAI

AI B

+=

Therefore

( )

( )( )

12

12

cos

sin 90A

c BAI

A B=

+

That is on using (2.2):

( )( )( )

( )

( )( )

( )

1

212

12

12

12

12

coscos

cos

cos 90

cos

sin

A

c BAI

A B

c B

C

c B

C

= +

=

=

But on using (4.7) c can be written as: 2 sinc R C=

giving

( )

( )

12

12

2 sin cos

sinA

R C BAI

C=

Using (2.11) gives

( ) ( ) ( )

( )

1 1 12 2 2

12

2 2sin cos cos

sinA

R C CAI

C

=

B

Hence ( ) ( )1 12 24 cos cosAAI R B C=

( ) ( )1 12 24 cos cosBBI R C A=

( ) ( )1 12 24 cos cosC R A B=

and similarly (8.13)

CI

a

bc

which give an alternative form of(8.12)

Further, from (8.12), as it is slightly easier we may proceed

( )( )

12 1

2

cosecsin

AA A

rAI r A

A= =


43/77

43

Thus, from (6.10) and (7.6) we have

( ) ( )( )

( )( )( )

( ) ( )( )

( )

( )

( )

( )

1A

AIs a s b s c

bcs s a s b s c bc

s a s b s c

bc s s a

s a

bc s

s a

bc

s a

s

=

=

=

=

=

Therefore:

or( )

A

bcAI

s a

s

=

( )B caBI

s b

s

=

( )C

ab

s c

s

=

similarly (8.14)

CI

a

bc

In a manner similar to the processes used to arrive at (8.11) we can also show that

(8.15)( ) ( ) ( )

1 1 1: : : :A B CAI BI CI

a s a b s b c s c=

Further, if we take respective products of(8.9) and (8.14) then we quickly get the

very nice results:

(8.16)

A

B

C

AI AI bc

BI BI ca

CI CI ab

=

=

=

a

bc


44/77

44

Y

XD

I

IA

A B

d .B C

II he distances , anA

II II

Figure 8.8

Consider the similar triangles andAID AI I Y, we have

A A

AI AD ID

II DX I= =

Y

That is

( )A A

s aAI rII AX AD I X Y

= = X

( )

( )A A

s aAI r

II s s a r I

= =

D

Hence,( )

A A

s aAI r

II a r

= =

r(8.17)

From the first two fractions we have

A

aII A

s a

=

B

bII

s b

=

C

cII C

s c

=

I

similarly (8.18)BI

I

Therefore, on using the values from the first forms in (8.10), the results in (8.18)

can be written in the manner

( )A

s aa abcII

s a s a

=

( )A

abc aII

s s a=

Hence

(8.19)( ) ( ) ( )

, ,A B Cabc a abc b abc c

II II IIs s a s s b s s c

= = =

a

bc

.


45/77

45

Squaring the value of AII from (8.19) gives

( )( )

2

A

abc aII

s s a

=

Therefore, from the last two fractions of(8.17) this becomes

( )2 A

A

abc r r II

s r

=

(8.20)

From (5.2) we have4

abc

R =

and from (6.1) that rs =

Hence, (8.21)4abc

Rrs

=

Therefore, from (8.20) and (8.21) we have that

( ) , , (8.22)( )2

4A AI I R r r= ( ) ( )2

4B BI I R r r= ( ) (2

4C CII R r r= )

Moreover, using the half-angle formula in (7.5) we see that (8.19) can be

rearranged thus:

( ) ( ) ( ) ( )

2

12

1

cosA

a bc bc aII a a

s s a s s a As s a

bc

= = =

Hence

(8.23)( ) ( ) ( )1 12 2

, ,cos cos cos

A B C

a bII II II

A B C= = =

12

.c

Therefore, on using the results of (4.7), (8.23) can be rewritten in a further,alternative form

(8.24)( ) ( ) ( )1 12 24 sin , 4 sin , 4 sin .A B CI I R A II R B II R C= = =12


46/77

46


47/77

47

9: Further Triangle Centres

The Orthocentre of any Triangle ABC

H

F

E

DB

A

C

The perpendiculars drawn from the vertices of a triangle ABC to the oppositesides are concurrent at a point called the Orthocentre, H.From Figure 9.1 let AD, BE and CF be the perpendiculars

on BC, CA and AB respectively, and H the Orthocentre;

then

( )

( )

tan

cos tan 90 ,

cos cot

DH BD HBD

AB B C

c B C

=

=

=

Using (4.7) ccan be written as:Figure 9.1

c R ,2 sin= C

giving 2 sin cos cotDH R C B C=

Hence2 cDH h=

(9.1)2 c cEH h= bc

F

aos cos AR B C

os os BR C A

2 cos cos CH R A B h=

Further,

sin

sin

2 sin sin

DA AB B

c B

R C B

=

=

=

That is2 sin sin

2 sin sin

2 sin sin

DA R B C

EB R C A

FC R A B

=

=

=

(9.2)

a

bc

Therefore, from (9.2) and (9.1) we have that

2 sin sin 2 cos cos

HA DA DH

R B C R B C

=

=

Using (2.3) this becomes

( )

( )

2 cos

2 cos 180

HA R B C

R A

= +

=

Hence:

2 cosHA R A=

2 cosHB R B= (9.3)2 cosHC R C=

a

bc


48/77

48

Notice that

( )

( )

( )

22

2 2

2 2

2 2 2

22

2 cos

4 cos

4 1 sin

4 4 sin

4 2 sin

HA R A

R A

R A

R R A

R R A

=

=

=

=

=

Using (4.7) acan be written as: 2 sina R A= ,

giving 2 24HA R a2=

Hence, we have the alternative forms of(9.3) as

(9.4)

2 2

2 2

2 2

4

4

4

HA R a

HB R b

HC R c

=

=

=

a

bc

Exercise: Establish(9.1)to(9.4)for an obtuse-angled triangle

Again using Figure 9.1, note the following simply found forms of h h ., andA B Ch

12

12

Area of triangle

A

ABC BC AD

a h

=

=

Therefore,2 2

, and similarly , .A Bh ha b

= =

2Ch

c

=

a

bc


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49

The Pedal Triangle of any Triangle ABCBy joining the feet of the perpendiculars from the vertices to the opposite sides, a

new triangle is formed called the Pedal triangle.We use Figure 9.2 and call the triangle DEF the pedal triangle of triangle ABC.

H

F

E

DB

A

C

Now, since then the quadrilateral90BFH BDH= =

BFHD is cyclic. Considering the right-angledtriangle AEB we have that

180 90

90

ABE FBH A

A

=

=

Next, consider the chord FHof

the circle through BFHD.

We have, using angles in the sameFigure 9.2

segment, that:

90FDH FBH A= =

Similarly, since , then the quadrilateral BHEC is also cyclic90CEH CDH = = Therefore, from Figure 9.3 we see that from the right-angled triangle AFC that

180 90

90

ACF ECH A

A

=

=

Next, consider the chord EHof the circle through CDHE. We have, using angles

in same segment that:

H

F

E

D

A

C

90ECH EDH A= =

Thus we have shown that .bisectsHD EDF

Similarly, and .bisectsHE FED bisectsHF DFE

The orthocentre, H, of triangle ABC is theincentre of its associated pedaltriangle.

Moreover,

B(9.5)

180 2 ;

180 2 ;

180 2 .

EDF A

FED B

DFE C

=

=

=

Further, since BC, CA and AB are perpendicular to HD, HE and HF respectively,

the sides of the fundamental triangle ABC are therefore the external bisectors of

the angles of the pedal triangle.

A, Band C are the excentres of the triangle DEF. (9.6)

Figure 9.3


50/77

50

H

F

E

DB

A

C

Figure 9.4

We also haveCDE BDF A

AEF CED B

BFD AFE C

= =

= =

= =

Hence, from using the Sine Rule (4.6) in the triangle AFE we have:

( ) ( )sin sinsin FEA EFAAEF FA EA

= =

The first two fractions lead tosin sin

cossin sin

cos

A B

EF AC A

A B

EF b A

=

=

Hence

sin cossin cos

sin sin

b A A bEF A A

B B= =

Therefore, from (4.7)

2 sin cosEF R A A= (9.7)

Therefore, using either (2.11) or (4.7), equation (9.7) can now be written in either

of two forms

( )sin 2EF R A= or cosEF a A= a

( )sin 2FD R B= or cosEF b B= (9.8)

( )sin 2DE R C= or cosDE c C= bc


51/77

51

The Circumcircle and the Pedal Triangle.Let 'R be the radius of the circle circumscribing the pedal triangle DEF

(Figure 9.2). Then, by the Sine Rule (4.7)

( )2 'sinEF R FDE= Therefore, by (9.8) and (9.5) we have

( ) ( )sin 2 2 'sin 180 2R A R A=

that is ( ) ( )sin 2 2 'sin 2R A R A= , by (2.2)

Hence; (9.9)12'R R=

That is, the radius of the circumcircle of the fundamental triangle ABC is twice

the radius of the circumcircle of the pedal triangle of the fundamental triangle

ABC

(Note: The Circumcircle of the pedal triangle of the fundamental triangle ABC i

actually theNine Point Circle of the fundamental triangle ABC)s

The Excentric Triangle.Returning briefly to the work of 8, which dealt with the Excircles of the

fundamental triangle ABC having centres , andA B CI I I and associated radii

, we make the following comparisons between these and those of thepedal triangle as discussed above.

, andA B Cr r r

I

IB

IA

ICA

B C

X

Figure 9.5


52/77

52

Recall from the construction of Incircle and Excircle that IC bisects andACB

AI C bisects the angle BCX .

Therefore,

( )

( )

1 12 2

12

12 180 90

A AICI ICB I CB

ACB XCB

ACB XCB

= +

= += +

= =

Similarly, BICI is also a right angle.

Hence, is a straight line to which IC is perpendicular

IA

I

B

A

C

A BI CI

B CI AI

C AI BI

is a straight line to which IA is perpendicular (9.10)is a straight line to which IB is perpendicular

Also, since AIand both bisect theAAI BAC

andB

, the three points all lie

on the same straight line. Similarly,

, and AA I I

CB I I C I I are also straight lines.

Hence, A B CI I I is a triangle, which is such that A, Band C are the feet of the

perpendiculars drawn from its vertices upon the opposite sides, and such that I is

the intersection of these perpendiculars. That is, triangle ABC is the pedal

triangle and I is the Orthocentre of the Excentric triangle A B CI I I .

We also note from Figure 9.6 that

( )12 180AI BC B= and ( )12 180AI CB C=

Therefore,

( ) ( )( )

1 12 2

12

12

180 90 90

90

ABI C B C

B C

A

=

= +

=

But as ABICI is a cyclic quadrilateral, then

Figure 9.6 1290BIC A= +

Similarly,

1 12 2

1 12 2

90 , 90 ,

90 , 90 .

B CCI A B AI B C

CIA B AIB C

= =

= + = +

Thus, triangles , ,A B CI BC I CA I AB are similar,

each with angles 1 12 290 , 90 , 90A B . (9.11)12 C

IA

X

IO

B C

Figure 9.7

From Figure 9.7, let X be the intersection of the

angle bisector through and the, and AA I I

circumcircle of triangle ABC.


53/77

53

Therefore, BX XC= as BAX CAX= . Moreover, 12BAX BCX A = ,

(from chord BX).

Therefore, ( )12XCI A C= +

but (12

)XIC A C= + when taken as the exterior angle to triangle AIC.

Hence, XC XI=

We also have 12AXCI B= as , and90AICI = ACX ABC B = , (from

chord AC). Therefore,12A

XI C B= , AXC X I=

Thus 12A AXB XC X I X I II= = = = .

Moreover, as AI BIC is a cyclic quadrilateral on AI I as diameter, the

circumference of the triangle AI BC passes through I.Similarly, the circumcircles of the triangles BI CA and CI AB also pass through I.

In the triangle ABI C, using the Sine Rule from (4.7), we have that

( )sin 1A

A

BI C

BC I=

IBIC, where 2 radius of AA II I

Therefore, from (9.11)( )12sin 90

A

aII

A=

Hence

( ) ( ) ( )1 12 2, ,

cos cos cosA B C

a bII II

A B C= = = (9.12)

a

bc12

,c

II

as met previously in (8.23)

IA

IB

IC

I

B

A

C

( )1 12 290 90B B B + = +

Consider now the Figure 9.8

Using (9.11) we see that C BBI I C

is a cyclic quadrilateral since

( ) ( )1 12 290 90

180

C B CI BC CI I

B B

+

= + +

=

1290 B

Moreover, as C AIB I I then C AI I

is a diameter of theC BBI I C

Therefore, in the triangle CBI C,

using the Sine Rule from (4.7),

we have that:( )sin 1C

B C

BI C

BC I=

I.

Figure 9.8


54/77

54

But 1290CBI C A= from (9.11), and from (9.10)90AI CI =

12C

BI C A = .

Therefore, we have

( )12sin 1

B C

A

a I=

I

Hence

(9.12)

( )

( )

( )

12

12

12

,sin

,sin

.sin

A B

B C

C A

cI I

C

aI I

A

bI I

B

=

=

=

a

bc

On using (4.7) these relations can easily be shown to take the alternative forms:

(9.13)( ) ( ) ( )1 12 24 cos , 4 cos , 4 cos .A B B C C AI I R C I I R A I I R B= = =12

From the Excentric triangle A B CI I I we have the simple area calculation:

( )( ) ( )

( )( ) ( )( ) ( )

12

1 1 12 2 2

sin

4 cos 4 cos sin 90

A B C C A A B C A BArea I I I I I I I I I I

12

R A R C A

=

=

Hence

(9.14)( ) ( ) ( )2 1 1 12 2 28 cos cos cosA B CArea I I I R A B C=

Notice that on using (7.5) we can rewrite (9.14) as

( ) ( ) ( )

( )( )( )

2

32

2 2 2

2

8

8

8

42

12

4

12

A B C

s s a s s b s s cArea I I I R

bc ca ab

s s a s b s cR

a b c

R s

abc

RR s

abc

R sabc

R

R s

=

=

=

=

=

=

Hence

(9.15)2A B CArea I I I Rs=

, from (5.2)


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55

10: Special Cevian Lengths

A Cevian is the name given to any line from a triangle vertex to its opposite sideans of any Triangle.The Centroid and Medi

Figure 10.1 the points D, E and F are midpoints of

gle ABC, usual

C

From page 3 of the Appendix we have

G

D

EF

B

A

C

In

the respective sides opposite the apexes A, B and C.

The lines , andAD BE CF are called the Mediansof the trian ly denoted

, , .AD m BE m CF m= = = A B

Figure 10.1

proved the following results:

2 2AG = 3 3, , .AD BG BE CG CF= = 23

The point G is called the Centroid of the triangle ABC.rom the Cosine Rule of (4.5) we have, from triangle ADC, in Figure 9.10

os .

That is

F

that

( ) ( ) ( ) ( )( )2 2 2

2 cAD AC CD AC CD C= +

( )

2

2 2

2 co2 2A

a a

m b b C

= +

s .

Hence2

2 2 cos4

A

am b ab C = + . (10.1)

However, a second application of the Cosine Rule in the whole triangle ABC gives

C. (10.2)

Hence, on equating (10.1) and (10.2) we find that

2 2 2 2 cosc b a ba= +

2 2 222

A

am c b =

2

22 2 22

2A

am b c= +

22 2 2

2B

bm c a= +

22 2 2

2C

cm a b= +

2

2

a

bc

Thus we have

(10.3)


56/77

56

On using the Cosine Rule once again the relations of(10.3) can be written:

os

os (10.4)

os

otice the elation derived from adding each of the results of

2 2 24 2Am b c bc= + + a

c A

c B

c C

2 2 24 2Bm c a ca= + +

2 2 2

4 2Cm a b ab= + +

simply determined r

bc

N

(10.3):

( )2 2 2 2 2 234A A Am m m a b c+ + = + + (10.5)

Figure 10.2 let

In ACADC = ; draw AL BC , then

( )

( )

( )

12

12

cot

cos

sin2 sin cos sin

2 sin sin

2sin cos sin

2sin sin

sin

2sin sin

cot cot

AC

DL BL BD

AL AL

c B a

c B

R C B R A

R C B

C B B C

B C

C B

B C

B C

= =

=

=

+=

=

=

hat isT

( )2cot cot cot sin cosec cosecAC B C C B B C =

Also, as 180AB AC = then

( )

( )

cot cot 1801

tan 180

1 tan180 tan

tan180 tan

1 0

0 tan

cot

AB AC

AC

AC

AC

AC

AC

=

=

+=

+=

=

Hence( )

( )

2cot cot cot sin cosec cosec

2cot cot cot sin cosec cosec

AC

AB

B C C B B

C B B C C B

=

=

C(10.6)

cb

mAK

L

G

DB C

Figure 10.2

A


57/77

57

To find the angles which makes withAm , drawAB CA DK AB and let

BDAB = and CDAC =

cot BAK

=

; then

( )

2

cotsin

12 cot

sinsin 1

2 cosin sin

sin2 cot

sin sin

sin2 cotsin sin2cot 2cot cot

a

AB KB c KB

KD KD KD KD

c B

B

cB

a B

CB

A B

CB

A B

A BBA B

B A B

= =

=

=

=

=

+

=

= +

t

Hencecot 2cot cot

cot 2cot cot

B A B

A C

= +

= +(10.7)

imilarly symmetric relations eS ian lines andBE CF.

C

xist for the other med

ng the results of(10.7) we have:Notice further, that on subtracti

cot cot cot cotB C = B C (10.8)

Hence, from (10.8), the identities of(10.6) can be written:

2cot cot cotAC B C

2cot cot cotAB C B

=

= (10.9)

evians Bisecting Angles InternallyC

c b

XB C

internally and divides the

ase BC into portions

AAIn Figure 10.3 the cevian AX bisects

b AB and AC

By the Sine Rule in triangle we haveAXB

( ) ( )12 ssin sin AXBA B= =

( )in

AB Ax c

(10.10)

riangle we haveand in t AXC

( 12sin

Figure 10.3

) ( ) ( )sinsin

AC A

AXCA C

x = =

b(10.11)


58/77

58

No AXB( ) ( ) ( )sin sin 180 sinAXC AXB = ting that

then we have on division of (10.11) and (10.10) that

ACx b=

ABx c(10.12)

Therefore, on using Theorem (3.2) we have

( ) ( )1 1AB ACAB AC AB ACx xx x x

b c b

+ += = =

+ +

Hence

x a

c( ) ( )11c b b c+

,AB ACac ab

x xc b c

= =+ +b

(10.13)

Also, if

bc

a

is the length of the cevian andAX AC AXC =

Therefore, ( ) (1 1 1 1sin sinc A + =

, then we have

Area AXB Area AXC Area ABC + =

) 12 2 2 2 2 sinb A bc A

Giving( )12

sin

sinA

bc A

b c A

=

+

Hence

( )122

cosAbc

Ab c

= +

(10.14)

hich on using (7.5) becomesw

( )2

As s a bc = (10.15)

b c+

2

2

AC

AB

AXAB B B

AXAC C C

= + +

= + +

otice also thatN

(10.16)

Clearly, the relationships of(10.12) to (10.16) can be re-written relative

to the other triangle apexes B and C by the usual symmetry rule


59/77

59

Cevians Bisecting Angles ExternallyIn Figure 10.4 the cevian bisects externally and divides theAX A

base BC extended, into portions ABx and ACx

c b

X 'XB

A

C

Figure 10.4By the Sine Rule in triangle ABX we have

( ) ( ) ( )12sin 90 sinsin AAB

A A B

x AX c

+ = =

That is:

( ) ( ) ( )12sin 90 sinsin AAB

A B

x AX c

+ = =

(10.17)

By the Sine Rule in triangle ACX we have

( ) ( ) ( )12sin 90 sin 180 sin AAC

A C

x AX b

= =

That is:

( ) ( ) ( )12sin 90 sinsin AAC

A A B

x AX b

+= =

(10.18)

Thus (10.17) gives

( ) ( ) ( )

1

2cos sinsin AAC

A A B

x AX b

+= = (10.19)

and (10.18) gives

( ) ( ) ( )12cos sinsin A

AB

A B

x AX c

= =

(10.20)

(10.19) (10.20) gives

1 1

11

AC

AB

x b

cx

=


60/77

60

Hence (10.21)AB

AC

c

x b

=

which is a similar relation to (10.12)

Therefore, on using Theorem (3.2) once again we have

( )

( )

1

1AB ACAB AC AB AC

x xx x x

c b b c b c b

+ = = =

+

( )

( )

1

1 c

x a

Hence

,AB ACa c ab

x xc b c b

= =

(10.22)

Also, if A is the length of the cevian AX and A AX C = , then we have

Area ABX Area ACX Area ABC =

Therefore,

( ) ( )

( ) ( )( )( ) ( )( )

1 1 1 12 2 2 2

1 12 2

1 12 2

sin 90 sin 90 sin

sin 90 sin 90 sin

cos cos sin

A A

A

A

c A A b A bc

c A b A bc

c A b A bc

+ + =

+ + =

+ =

12

A

A

A

Giving( )12

sin

cosA

bc A

c b A

=

Hence

( 122

sinAbc

Ac b

= (10.23)

)

which on using (7.6) becomes

( )(2

A (10.24))bc s b s cc b

=

Notice also that

(10.25)

90

2180 90

2

A

A

AC

AB

= +

= + +

a

bc

Clearly, the relationships of(10.21) to (10.25) can be re-written relative

to the other triangle apexes B and C by the usual symmetry rule

One further observation is that

AC AC

ab abX X x x

b c c b = + = +

+

Hence (10.26)2 2

2abcX X

c b =


61/77

61

DB C

AApollonius TheoremIn Figure 10.5 the point D bisects the base BC

of the triangle ABC.

Apollonius Theorem states that:

( ) ( ) ( ) ( )2 2 2 2

Proof: Let

2 2AB AC AD BD+ = +

ADB = ; therefore 180ADC = .

es

os

Therefore, using the Cosine Rule giv

Figure 10.5

( ) ( ) (AB AD D= + ) ( )( )2 2 2

2 cB AD DB (10.27)

) ( ) ( ) ( ) ( )( ) (2 2 2 2 cos 180AC AD DC AD DC= +

That is )os( ) ( ) ( ) ( )( ) (2 2 2

2 cAC AD DC AD DC = + (10.28)

re, by the addition of(10.27) and (10.28), and sinceTherefo DC= we have:BD

( ) ( ) ( ) ( )2 2AB AC AD BD+ = + 2 2 2 2

Generalisation of Apollonius Theorem Ste oremarts The

ivided in the ratio

In Figure 10.6 the base BC of the triangle ABC has been

d

:BD DC :m n= (10.29)

by the foot D of the general cevian AD

0.27) by and (10.28)

y to give

F

b

urther, we have on multiplication of (1 n

( ) ( ) ( ) ( )( )2 2 2

2 cosn AB n AD n DB n D DB = +

B C

A

D

Figure 10.6

m

and

A

) (10.31)

(10.30)

os( ) ( ) ( ) ( ) ( ) (2 2 2

2 cm AC m AD m DC m AD DC = + +

From (10.29),(10.31) can be written

)os( ) ( )m AC m AD m= + ( ) ( )( ) (2 2 2

2 cDC n AD BD + (10.32)

ence, adding (10.30) and (10.32) gives us the general form

2(10.33)

hich redu

H

( ) ( ) ( )( ) ( )2 2 2 2

n AB m AC m n AD n BD m+ = + + + ( )DC

w midpoint of ces to Apollonius form for m n= , that is forD being the

BC.


62/77

62

Figure 10.8

by fi

the Figure 10.5 in the following manner.

Stewarts Theorem can then be shown in a form

more easily recalled, namely:

(10.34)

In the same manner we can extend the results of(10.16) for a general cevian

rather than the angle bisector of that section.

Consider Figure 10.8 in which we set

Figure 10.7

c

ba d

m n

This is perhaps most easily remembered rst re-labelling

( )2 2 2a n b m c d mn+ = +

: :BD DC m= n

c b

nmB

A

CD

That ism BD BD AD

n DC AD DC

= =

Therefore, on using the Sine Rule we have

( )

( )( )( )

( )

sin 180 180sin

sinsin 180

AB CB

CB AB

m BD AD

n AD DC

= =

,

( )

( )

( )

( )

( )

( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ){ } ( )( ) ( ) ( ) ( ){ } ( )

( ) ( ) ( )

( ) ( ) ( )

sin sin

sin sin

sin sin cos cos sin

sin sin cos cos sin

sin cos cos sin / sin

sin cos cos sin / sin

sin cot cos

cos cot sin

B AB C

C B AB

B AB C AB

C B AB B

AB C AB C C

B AB B AB

AB C AB

AB B AB

m

n

C

AB

B

=

+

=

+

=

+

=

+

That is

Hence,

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

cos cot sin sin cot cos

cos sin cot cot sin ,

AB B AB AB C AB

AB AB C B AB

m m n n

m n n m

+ =

+ =

giving: ( ) ( ) ( ) ( )cot cot cotAB C Bm n n m + = (10.35)

Similarly, using the identity ( ) ( )cot 180 cot gives

( ) ( ) ( ) ( )cot cot cotAC B Cm n m n + = (10.36)

Notice that (10.35) and (10.36) reduce to (10.9) for m n= , that is forD being the

midpoint ofB.


63/77

63

11: Problems

Problem 1. Prove:4r r r r R+ + =A B C

Problem 2. Prove:1 1 1

A B Cr r r

1

r+ + =

Problem 3. Prove:1 1 1

B Ch h+ + =

1

r

Ah

ove:Problem 4. Pr2

A B Cr r r rs= 2 A B Cr r r r =

Problem 5. Prove:

( ) ( ) ( )

( ) ( ) ( )

1 1 12 2 2

1 1 12 2 2

4 cos cos cos

4 cos cos cos

s R A B C

Rr A B

=

=

C

6. Prove:

C

rove:

Problem

cos cos cos 4 sin sin sina A b B c C R A B+ + =

Problem 7. P2

A B B C C Ar r r r r r s+ + =

Problem 8. Prove:

( )21 1 1 1

4A B C A B C

abcs

r r r r r r R r + + =

+ + +

Problem 9. Prove:

( ) ( )2 1 12 24 cos cotB Cr r R A a A+ =

10. Prove:Problem

)r r+ ( )( ) ( ) (4A B B C C A A B B C C Ar r r r r r R r r r r + + + = +

Problem 11. Prove:

( ) ( )( ) 24A B Cr r r r r r R r =


64/77

64

Problem 12. Prove:3

2 2 2 2 2

1 1 1 1 1 1 1 1 1 1 1 1 64 4

C C A A B C

R R

r r r r r r r r r a b c r s

+ + + = =

A B Br r r

roblem 13. Prove:P

( ) )( ) ( )( ( ) ( ) 2

1 1 1

s a s b s b s c s c s a r

1+ + =

Problem 14. Prove:

( )22 2 2 2 2 24 1b c b c a 6= + +

Problem 15. Prove:

( ) ( )2 24 sin 2 sin 2b A a = + B

Problem 16. Prove:

)

( ) ( ( )2 1 1 12 2 2tan tan tans A B = C

Problem 17. Prove:22 sin sin sinR A B = C

Problem 18. Prove:

( )sin sin sinRr A B C = + +

Problem 19. Prove:

( )

2 2 sin sin

2 sin

a b A B

A B

=

Problem 20. Prove:

( ) ( ) ( )( )3 2 2 232 sin 2 sin 2 sin 2a b c A B C = + +

Problem 21. Prove:

( )

2 2 2

4cot cot cot

a b cA B C

+ + =+ +

Problem 22. Prove:

rr( ) ( )( )A B C A B C AB C

B CA B B C C A

r r r r r r a r

s rr r r r r r

+ += = = +

+ +r

r

Problem 23. Prove:

( ) ( )2 A Ba r r r r = + C


65/77

65

Problem 24. Prove:

( )( ) ( )

12sin

A

A B A C

rA

r r r r =

+ +

Problem 25. Prove:

( )( )

2sin A A B B C C A

A B A C

r r r r r r r A

r r r r

+ +=

+ +

Problem 26. Prove:

(

( ) ( ) )3 2 2 21 1 12 2 2cot cot cotA B Cr r r r A B C =

Problem 27. Prove:2 2

2 2 2 2 2

1 1 1 1

A B C

a b c

r r r r

2+ ++ + + =

Problem 28. Prove:2

Problem 29. Prove:

2 2 2 2 2 2 216A B Cr r r r R a b c+ + + =

A B Cab r r rr = +

Problem 30. Prove:

1 1 1 12ab bc ca Rr

+ + =

Problem 31. Prove:1 1

2CA B

rr r

bc ca ab r R+ + =

Problem 32. Prove:

0A B C

b c c a a b

r r r

+ + =

Problem 33. Prove:

( ) ( ) ( )A B C B C A C A Ba rr r r b rr r r c rr r r + = + = +

Problem 34. Prove:

A B B C C A A B Cr r r r r r rr rr rr ab bc ca+ + + + + = + +

Problem 35. Prove:

(

( ) ( ) ( )

2 2 28 2

A B C

I I I I I I R R+ + = )r


66/77

66

Problem 36. Prove:216A B CI I I I I I R r =

Problem 37. Prove:

( ) ( )2

4B C B CI I R r= + r

Problem 38. Prove:

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2

A B C B C A C A

2

BI I I I I I I I I I I I+ = + = +

Problem 39. Prove:

sin sin sinA B C B C A C A BI I I I I I I I I I I I

A B

= =

C

Problem 40.

radii of three circles which touch one another externally

gents at their points of contact meet in a point; prove that the

his point from eitherof their points of contact is

, anda b c are the

and the tan

distance of t

abc

a b c+ +


67/77

Appendix: Concurrences of Straight Lines in a Triangle


68/77

68


69/77

69

69

ircumcentre:The perpendiculars drawn to the sides of a triangle from their

midpoints are concurrent

Let X, Y and Z be the midpoints of the side

triangle ABC.

From Zand Ydraw perpendiculars to

meeting at O. Join OX.

Thus, it is required to show that OX isperpendicular to BCJoin OA, OB, OC.

Proof:

C

s of the

AB, AC

Z

O

X

Y

B

A

C

Because YO bisects AC at right angles, it is therefore the locus of points that are

equidistant from A and C.

Therefore, OA OC=

Again, because ZO bisects AB at right angles, it is therefore the locus of points

that are equidistant from A and B.

Therefore, OA OB=

Hence: OB OC=

Therefore, O is on the locus line of points equidistant from B and C; that is

Hence, the perpendiculars from the midpoints of the sides of the triangle ABCmeet at O.

is perpendicular toOX BC


70/77

70

Incentre:

isect ,

The bisectors of the angles of a triangle are concurrent

O. JoinABC BCA

B by straight lines that meet at

red to show that AO bisects

.

AO.

BAC .e R

P

Q

O

B

A

C

Thus, it is requi

From O draw OP, OQ, OR perpendicular to th

sides of the triangle ABC

Proof:

Because BO bisects it is

t to the locus

of points that are equidistant

A and BC.

BAC

therefore equivalen

from B

Therefore, OP OR=

Similarly, CO is the locus of points that are equidistant from CB and CA.

herefore OP OQ= T

Hence OR OQ=

Therefor

is, OA i

e, O is on the locus of points that are equidistant from AB and AC; that

s the bisector of the

ence the bisectors of the three angles of the triangle ABC meet at O.

BAC.

H


71/77

71

Centroid:The medians of a triangle are concurrent

Let BY Z o of the medians of the triangle ABC, and leand C be twt O. Join AO and produce it to meet

hus, it is required to show that AX

hrough C draw CK, parallel to BY

CK at K. Join BK.

t them intersectBC in X.

is the remaining median

. Produce AX

O

K

Y

X

Z

B

A

C

a

Tof triangle ABC.T

to meet

Proof:

Because Y is the middle point ofAC,

to CK, therefore O

the middle point ofAK.

triangles )

e have

parallel to BK, that is, OC is parallel to BK.

herefore, the figure BKCO is a parallelogram. But the diagonals of aogram bisect one another; therefore X s the middle point ofBC.

ence, the medians of triangle ABC meet at O.

orollary:

and YO is parallel

is

AYO ACK (By similar

In the triangle ABK, since Z and O are the middle points ofAB and AK w

that ZO is

Tparallel i

That is, AX is also a median of the triangle ABC.

H

(O is usually referred to as the Centroid of the triangle)C

The three medians of a triangle cut one another at a point of trisection, the greatersegment in each being towards the angular point.

We have seen above that AO OK= ,

and that OX is half ofOK;

therefore OX is half ofOA:

that is OX is one third ofAX.

Similarly, OY is one third ofBY,

and OZ is one third ofCZ.


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Orthocentre:The perpen ngle to the

opposite sides are concurrent

m A and B to their opposite

des; and let them intersect at O. Join CO;and produce it to meet AB at F.

d to show that CF is

diculars drawn from the vertices of a tria

Let AD and BE be the perpendiculars drawn fro

si

Thus, it is require perpendicular to AB.F

D

E

O

B

A

C

Proof:

Join DE.

ecause and are both

ght-angles, we have:

cyclic

um to )

, vertically opposite angles.

ave:

, in same segment.

Therefore, we ha

(angles in triangle sum to )

CF is perpendicular to

Hence the three perpendiculars AD, BEand CF meet at the point O.

OEC ODCThen, b

ri

Points O,E,C,D are concyclic.

(Opposite angles in a

180quadrilateral s

Therefore, DEC DOC= , in same segment,E

O

D C

DEC FOA=

Similarly, as AEB and ADB are both right-angles, we

h

Points A,E,D,B are concyclic.

(Opposite angles in a cyclicquadrilateral sum to180 )

Therefore,

DEB DAB=

veO

D

E

F

B

A

C DE=

90

FOA FAO DEC DAB

C DEB

+ +

+

=

Therefore, 90AFO = 180

That is, AB.


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Bibliography


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In order of usage:

Sixth Form Trigonometry. W. A. C. Smith. James Nisbet & Co. Ltd. [1956].

Modern Geometry. C. V. Durell. Macmillan & Co. Ltd. [1957].

Plane Trigonometry (Part 1). S. L. Loney. Macmillan & Co. Ltd. [1967].

A School Geometry (Parts I VI) . Hall & Stevens. Macmillan & Co. Ltd. [1944].

Advanced Euclidean Geometry: An Elementary Treatise on the Geometry of the

Triangle and the Circle. Roger A. Johnson. Dover Publications Inc. [1929].

College Geometry: An Introduction to the Modern Geometry of the Triangle and

the Circle. N. Altshiller Court. Barnes & Noble Inc. [1952].


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elementary triangle geometry (mark dabbs)

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