elementary_triangle_goemetry1.pdf

The Circumcircle and the Pedal Triangle.. 55The Centroid and Medians of any Triangle.. Motivating ProblemP C B A The motivation for this work came from an open question to a class to fin

Elementary Triangle Geometry

Mark Dabbs

The Mathematical Association Conference

University of York, U.K Spring 2004

Version 1.1 April 2004 (www.mfdabbs.com)

Contents

Area Formulae in terms of a, b and c

Sin A, sin B and sin C in terms of a, b and c

Cos A, cos B and cos C in terms of a, b and c

Tan A, tan B and tan C in terms of a, b and c

Incircle

Excircles

Heron’s Area Formula

Sin A , sin B and sin C in terms of , and a b c

Relationship between and r R

Relationships between , ,r r r A B C and R

The distances A ,I BI and C I

The distances A ,I BI A B and C I C

The distances II II A, B and II C.

The Orthocentre of any Triangle ABC The Pedal Triangle of any Triangle ABC The Circumcircle and the Pedal Triangle

The Excentric Triangle

§10: Special Cevian Lengths 55

The Centroid and Medians of any Triangle

Cevians Bisecting Angles Internally

Cevians Bisecting Angles Externally

Motivating Problem

P C

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

base height,

bh

where b AC= and h=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 BPC we have:

sinsin

PB C BC h C a

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

Note the following Pythagorean relations within the two triangles CBP and ABP

4

c a b a b c h

b

Now it’s 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 insist that such symmetry exist in these other three factors

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

c a b a b c a

c a b a b c b

a b c a b c c

− + ≡ + + −+ − ≡ + + −+ − ≡ + + −

) ) )

b

This is then easily factorised to give:

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

inexhaustible in its properties!”

(A L Crelle, 1821)

§1: Basic Trigonometrical Formulae

sin

tancos

sin1sec

cos1cot

tan

θ

θθ

θθ

§2: Further Trigonometrical Formulae

t t

Therefore, x c= cos ,A y a= cosC

Hence b x y c= + ≡ cosA a+ cosC

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

thus:

cos coscos coscos cos

The formulae of (4.1) are known as the Projection Formulae

If we now multiply the equations of (4.1) by a, b and c, respectively, we have:

Equation (4.5) is known as the Cosine Rule for triangles

Symmetry yields the other forms:

2 2 2 2 cos and 2 2 2 2 cos

Therefore, h c= sin or A h a= sinC

Hence, csinA a= sinC h≡

However, the initial orientation of the triangle ABC was arbitrary

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 Circumcircle of the triangle ABC)

P

P

O O

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( ) sin( )

From Figure 4.3 APC=180 − B (Cyclic Quadrilateral)

Therefore, sin( APC)=sin 180( −B)≡sinB

Hence, as for Figure 4.2 we have

Therefore, from (4.6) we have:

−

( ) ( )

2 .tan

1 2

1 2

tantan(90 )

Equation (4.10) is known as the Tangent Rule for triangles

Symmetry yields the other forms:

§5: Other Triangle Formulae

b A

B

C

Area Formulae in terms of a, b and c

Figure 5.1 The area of triangle ABC is found from

1 2 1 2

, where sin ,sin

b c

1

2

sinsinsin

a

A

∆ =Hence,

sin sin sin sin sin sin

Sin A in terms of a, b and c, etc

4141

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

a

c

b

From which, the Sine Rule can be deduced, since

2sin

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

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

Sin A, sin B, sin C in terms of a, b and c

2∆

Cos A, cos B, cos C in terms of a, b and c

From (4.5) we simply rearrange to yield

Tan A, tan B, tan C in terms of a, b and c

From (1.1), (5.5) and (5.8) we have

tanA 2 42 2, tanB 2 42 2, tanC 2 2 2 (5.9)

§6: Associated Circles

E F

Let I be the Incentre of the

triangle ABC, obtained by

bisecting the interior angles of

the triangle ABC Then

,

ID IE= =IF≡r

(ID IE, and IF are

where r is the radius of the incircle

the⊥'sfrom to the respective sides)I

Area of triangle ABC Areas of triangles BIC CIA AIB

Tangents to a circle from

a single point have equal lengths

From triangle ABC we see that

( ) ( ) ( )

1 2

1 2

1 2

tanntan

r A

s a r B

s b r C

1 2 1 2 1 2

cotcotcot

D I

B

C A

a

b c

( ) ( ) ( ) ( )

sin 90sin sin

1 2

cossin sin

1 2

sin sincos

r a

A

Together, we have

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

1 2

1 2

1 2

sin sincossin sincossin sincos

1 2

Excircles

A' D

Let I be the centre of a circle, opposite A

angle A, of radius obtained by bisecting

the angles B and C externally and A

internally

A

r A r

Let I be the centre of a circle, opposite A

angle A, of radius obtained by bisecting

the angles B and C externally and A

A A

Hence for s, the semi perimeter

Hence for s, the semi perimeter

s b r

Tangents to a circle from

a single point have equal lengths

Tangents to a circle from

a single point have equal lengths

a

b c

'

A

Figure 6.3

A' D

A A

Therefore:

1 2

The three distances x A =s x, B =sand x C =s

are displayed below

Figure 6.4Figure 6.5

Notice also from triangle AI F A ' and from (6.11) that

( )1 2

'tan

1 2 1 2 1 2

tantantan

A B C

Therefore, we can further show that

( ) ( ) ( )

1 2 1 2 1 2

cot , cot

A B C

tangency of the excircle with side a, and where the symbol is the positive

difference between its two arguments Namely:

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

is also the midpoint of BC

'

Heron’s Area Formula

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

Moreover, since IDBand I D B A '

§7: Further Triangle Formulae

Figure 7.1

Y

D I

I A

X A

C

B

( )1 ( )1 ( )1

Tan A , tan B and tan C in terms of , and a b c

From Figure 7.1 we have ( )1

tan s b s c s b s c A

1 2

1 2

1 2

tan

tan

tan

s b s c A

s s a

s c s a B

s s b

s a s b C

( )1 ( )1 ( )1

Cos A , cos B and cos C in terms of , and a b c

From Figure 7.1 we have 1( )

1 2 1

cos

A

AX AD A

(7.5)

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

1 2

1 2

1 2

coscoscos

s s a A

bc

s s b B

ca

s s c C

( )1 ( )1 ( )1

Sin A , sin B and sin C in terms of , and a b c

Using (1.1), (7.1) and (7.5) we have that

(7.6)

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

1 2

1 2

1 2

sinsinsin

s b s c A

bc

s c s a B

ca

s a s b C

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

angles, A, B and C are those of a triangle

§8: Further Triangle Relationships

Relationship between r and R

sincos

sincos

1

2 1

2

1 2

sin

sincos

sin sincos

2 sin sin sin

Relationships between , ,r r r A B C and R

coscos

coscos

1

2 1

2

1 2

cos

coscos

cos coscos

2 sin cos cos

X D

I

I A

C

The distances A ,I BI and C I

From Figure 8.3 notice that

=

Figure 8.3Therefore,

( ) ( ( )

1 2 1 2 1 2

coseccoseccosec

X D

I

I A

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

angles from Figure 8.4

and so on using the Sine Rule of (4.6)

we have from triangle AIB

sinsin

sinsin 180

1 2 1 2 1 2 1 2 1 2 1 2

sinsinsinsin 90sincos

( ) ( )

1 2 1 2

2 sin sincos

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

easier we may proceed ( )1 ( )

s

∆

a

b c

a

a b

b c c

X D

I A

=Therefore,

( ) ( ) ( )

1 2 1 2 1 2

coseccoseccosec

D

I

I A

B C

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

sin 90sin

cossin 90

1 2 1 2 1 2 1 2 1 2 1 2

coscoscoscos 90cossin

1 2 1 2

2 sin cossin

1 2

which give an alternative form of (8.12)

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

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

s a s

A

bc AI

s a s

s b s

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

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

X D

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

can be written in the manner

−

Squaring the value of II A from (8.19) gives

( )2 ( )

A

abc a II

∆ =and from (6.1) that ∆ =rs

Rr s

§9: Further Triangle Centres

The Orthocentre of any Triangle ABC

H

F

E

D B

A

C

The perpendiculars drawn from the vertices of a triangle ABC to the opposite

sides 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;

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

2 2

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 A, B and h C

1 2 1 2

The Pedal Triangle of any Triangle ABC

By 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

D B

A

C

Now, since BFH = BDH =90 then the quadrilateral

BFHD is cyclic Considering the right-angled

triangle AEB we have that

Next, consider the chord FH of

the circle through BFHD

We have, using angles in the same

Figure 9.2segment, that:

FDH = FBH =90 A

Similarly, since CEH = CDH =90 , then the quadrilateral BHEC is also cyclic Therefore, from Figure 9.3 we see that from the right-angled triangle AFC that

180 9090

A

= − Next, consider the chord EH of the circle through CDHE We have, using angles

in same segment that:

Thus we have shown that HD bisects EDF

Similarly, HE bisects FED and HF bisects DFE

⇒ The orthocentre, H, of triangle ABC is the

incentre of its associated pedal triangle

⇒ A, B and C are the excentres of the triangle DEF (9.6)

Figure 9.3

F

E

D B

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)

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 the Nine 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 I I A, B and I C and associated radii

, we make the following comparisons between these and those of the

pedal triangle as discussed above

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

ICI ICB I CB

ACB XCB ACB XCB

Similarly, ICI B is also a right angle.

Hence, is a straight line to which IC is perpendicular

I A

I B

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

is a straight line to which IB is perpendicular

Also, since AI and AI A both bisect the BAC

and

B

, the three points all lie

on the same straight line Similarly,

, and A

C

B I I C I I are also straight lines

Hence, I I I A B C is a triangle, which is such that A, B and 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 I I I A B C

We also note from Figure 9.6 that

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

each with angles 1 1

90 − A, 90 − B, 90 −1 (9.11)

2C

I A X

I O

Figure 9.7

From Figure 9.7, let X be the intersection of the

angle bisector through A I, and I A and the

circumcircle of triangle ABC

Therefore, BX = XC as BAX = CAX Moreover, 1

2

BAX ≡ BCX = A, (from chord BX)

circumference of the triangleI BC passes through A I

Similarly, the circumcircles of the triangles I CA and B I AB C also pass through I

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

A

=

−Hence

Consider now the Figure 9.8

Using (9.11) we see that BI I C C B

is a cyclic quadrilateral since

90 − B

Moreover, as IB⊥I I C A then I I C A

is a diameter of the BI I C C B

Therefore, in the triangle BI C C ,

using the Sine Rule from (4.7),

we have that: sin( C ) 1

B C

BI C

1 2

1 2

1 2

,sin

,sin

.sin

I I

A b

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

2 2 2 2

888

42

12

412

A B C

s s a s s b s s c Area I I I R

s s a s b s c R

a b c

R s abc R

abc

abc R

(9.15) 2

A B C

Area I I I = Rs

, from (5.2)

§10: Special Cevian Lengths

A Cevian is the name given to any line from a triangle vertex to its opposite side

ans of any Triangle

The Centroid and Medi

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

gle ABC, usual

B

A

C

In

the respective sides opposite the apexes A, B and C

The lines AD BE, and CF are called the Medians

of the trian ly denoted

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

Figure 10.1proved the following results:

AG=23AD BG, =32BE CG, = 2CF

3

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

Thus we have

(10.3)

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

1 2

cot

cossin

2 sin cos sin

2 sin sin2sin cos sin2sin sinsin

2sin sincot cot

1 tan180 tantan180 tan

1 0

0 tancot

AC AC AC

AC AC

θθθ

θθ

=

−+

=

−+

=

−

= −Hence

θθ

To find the angles which m A makes with AB CA, draw DK ⊥ AB and let

sin sinsin

sin sin2cot 2cot cot

a

AB KB c KB

KD KD KD KD c

B B

imilarly symmetric relations e

C

xist for the other med

ng the results of (10.7) we have:

Notice further, that on subtracti

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

2 cot AC cot B cot C

2 cot AB cot C cot B

In Figure 10.3 the cevian AX bisects

AXC

Noting that sin( AXC)≡sin 180( − AXB)=sin( AXB)

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

a

δ is the length of the cevian AX and ϕAC = AXC

Therefore, 1cδsin( )1A +1 sin(1 =

sinsin

A

δ =  + Hence

( )1 2

2cos

AC

AB

A XAB B B

A XAC C C

ϕϕ

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