Na geometrię, cóż więcej dodać, elementarna geometria trójkąta.

1. Elementary Triangle Geometry
Mark Dabbs
The Mathematical Association Conference
University of York, U.K
Spring 2004
Version 1.1 April 2004
(www.mfdabbs.com)

2. 2

3. 3
Contents
Motivating Problem. 5
§1: Basic Trigonometrical Formulae. 11
§2: Further Trigonometrical Formulae. 13
§3: Ratio Theorems. 15
Theorem (3.1).
Theorem (3.2).
§4: Basic Triangle Formulae. 17
Cosine Rule.
Sine Rule and Cicumcircle.
Tangent Rule.
§5: Other Triangle Formulae. 21
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.
§6: Associated Circles. 25
Incircle.
Excircles.
Heron’s Area Formula.
§7: Further Triangle Formulae. 33
Tan ( 1 A ) , tan ( 1 B ) and tan ( 1 C )
2 2 2 in terms of a , b and c.
Cos ( 1 A ) , cos ( 1 B ) and cos ( 1 C ) in terms of a , b and c.
2 2 2
Sin ( 1 A ) , sin ( 1 B ) and sin ( 1 C ) in terms of a , b and c.
2 2 2
§8: Further Triangle Relationships. 37
Relationship between r and R.
Relationships between rA , rB , rC and R.
The distances AI , BI and CI .
The distances AI A , BI B and CI C .
The distances II A , II B and II C .
§9: Further Triangle Centres. 47
The Orthocentre of any Triangle ABC .
The Pedal Triangle of any Triangle ABC.
The Circumcircle and the Pedal Triangle.
The Excentric Triangle.

4. 4
§10: Special Cevian Lengths. 55
The Centroid and Medians of any Triangle.
Cevians Bisecting Angles Internally.
Cevians Bisecting Angles Externally.
Apollonius’ Theorem.
A Generalisation of Apollonius’ Theorem – Stewart’s Theorem
§11: Problems. 63
Appendix: Concurrences of Straight Lines in a Triangle. 67
Circumcentre.
Incentre.
Centroid.
Orthocentre.
Bibliography. 73

5. 5
Motivating Problem
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.
B
A typical diagram is shown in Figure MP.1
a h c
P
C A
x b-x
Figure MP.1
The area of the triangle, ∆ , is seen to be
∆ = 1 ( base ) ( ⊥ height )
2
(MP.1)
∆ = 1 bh,
2
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:
PB
sin C =
BC
h
sin C =
a
Therefore, h = a sin C (MP.2)
From (MP.1) and (MP.2) we have the general area formula
∆ = 1 ab sin C
2 (MP.3)

6. 6
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.
a 2 = x 2 + h2 (MP.4)
c 2 = ( b − x ) + h2 .
2
and (MP.5)
Eliminating h from (MP.4) and (MP.5) gives:
a2 − x2 = c2 − (b − x )
2
a2 − c2 = x2 − (b − x ) .
2
That is:
Thus, by the difference of two squares formula we have
a 2 − c 2 = ( x − b − x )( x + b − x )
a 2 − c 2 = ( x − b + x )( x + b − x )
a 2 − c 2 = ( 2 x − b )( b )
Hence,
a2 + b2 − c2
x= . (MP.6)
2b
Substituting (MP.6) into (MP.4) gives
2
 a2 + b2 − c2 
h = a −
2 2
 .
 2b 
That is
4 a 2b 2 − ( a 2 + b 2 − c 2 )
2
h =
2
(MP.7)
4b 2
Once again, by the difference of two squares formula we have the alternative form
of (MP.7):
h 2
=
( 2ab − a 2
)(
+ b2 − c 2 2ab + a 2 + b 2 − c 2 ),
4b 2
=
( 2ab − a 2
− b 2 + c 2 )( 2ab + a 2 + b2 − c 2 )
,
4b 2
=
(c 2
− a 2 + 2ab − b 2 )( a 2 + 2ab + b2 − c 2 )
,
4b 2

7. 7
That is:
h 2
=
(c 2
− a 2 − 2ab + b2 )( a 2
+ 2ab + b 2 − c 2 ),
4b 2
which, on factorising gives:
h 2
=
(c 2
− (a − b)
2
) (( a + b) 2
− c2 ).
4b 2
Therefore, on using the difference of two squares formula again we have:
h2 =
( c − a − b )( c + a − b )( a + b − c )( a + b + c )
4b 2
( c − a + b )( c + a − b )( a + b − c )( a + b + c )
h2 = . (MP.8)
4b 2
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
following manner:
( c − a + b ) ≡ ( a + b + c − 2a )
( c + a − b ) ≡ ( a + b + c − 2b ) (MP.9)
( a + b − c ) ≡ ( a + b + c − 2c )
Realising that ( a + b + c ) is just the perimeter of the original triangle ABC, say p
gives (MP.8) as:
( p − 2a )( p − 2b )( p − 2c )( p )
h2 = . (MP.10)
4b 2
However, if we then let the new variable, s, be defined as the semi-perimeter then
(MP.10) is re-written
( 2s − 2a )( 2s − 2b )( 2s − 2c )( 2s )
h2 = .
4b 2

8. 8
This is then easily factorised to give:
16 ( s − a )( s − b )( s − c )( s )
h2 =
4b 2
or
4s ( s − a )( s − b )( s − c )
h2 = .
b2
2 s ( s − a )( s − b )( s − c )
Hence, h= (MP.11)
b
Finally then, substituting (MP.11) back into (MP.1) gives
 2 s ( s − a )( s − b )( s − c ) 
∆ = 1 b× 
2
 b 
 
or
∆ = s ( s − a )( s − b )( s − c ) (MP.12)
Which is the familiar result of Heron of Alexandria (First Century A.D)

9. 9
“What a marvel that so simple a figure as the triangle is so
inexhaustible in its properties!”
(A. L. Crelle, 1821)

10. 10

11. 11
§1: Basic Trigonometrical Formulae
sin θ
= tan θ (1.1)
cos θ
sin 2 θ + cos 2 θ = 1 (1.2)
θ 0 30 45 60 90
1 2 3
sin θ 0 1
2 2 2
(1.3)
3 2 1
cos θ 1 0
2 2 2
1
tan θ 0 1 3 ∞
3
1
cosec θ =
sin θ
1
sec θ = (1.4)
cos θ
1
cot θ =
tan θ
tan 2 θ + 1 = sec 2 θ (1.5)
1 + cot 2 θ = cosec 2 θ (1.6)

12. 12

13. 13
§2: Further Trigonometrical Formulae
sin ( A + B ) = sin A cos B + cos A sin B (2.1)
sin ( A − B ) = sin A cos B − cos A sin B (2.2)
cos ( A + B ) = cos A cos B − sin A sin B (2.3)
cos ( A − B ) = cos A cos B + sin A sin B (2.4)
tan A + tan B
tan ( A + B ) = (2.5)
1 − tan A tan B
tan A − tan B
tan ( A − B ) = (2.6)
1 + tan A tan B
P+Q   P−Q 
sin P + sin Q = 2 sin 
  cos   (2.7)
 2   2 
P +Q   P −Q 
sin P − sin Q = 2 cos 
  sin   (2.8)
 2   2 
P+Q  P−Q 
cos P + cos Q = 2 cos 
  cos   (2.9)
 2   2 
P +Q   P −Q 
cos P − cos Q = − 2 sin 
  sin   (2.10)
 2   2 
sin ( 2 A ) = 2sin A cos A (2.11)
cos ( 2 A ) = cos2 A − sin 2 A = 2cos2 A − 1 = 1 − 2sin 2 A (2.12)
sin 2 A = 1
2 (1 − cos ( 2 A) ) (2.13)
cos 2 A = 1
2 (1 + cos ( 2 A) ) (2.14)
Notice further, if we define t = tan 1 θ then it can be shown that
2
2t 1 − t2 2t
sin θ = , cosθ = and tan θ = . (2.15)
1 + t2 1+ t 2
1 − t2

14. 14

15. 15
§3: Ratio Theorems
Theorem (3.1)
p a λ p + µ q λ a + µb
If we have that = then = , for any numbers λ , µ , m and n.
q b mp + nq ma + nb
Proof:
p a
Let = ≡t
q b
⇒ p = qt , a = bt .
λ p + µq λ ( qt ) + µ q λt + µ
Therefore, = =
mp + nq m ( qt ) + nq mt + n
and
λ a + µb λ ( bt ) + µ q λt + µ
= = □
ma + nb m ( bt ) + nq mt + n
Theorem (3.2)
p a λ p + µa p a
If we have that = then = = , for any numbers λ and µ .
q b λ q + µb q b
Proof:
p a
Let = ≡t
q b
⇒ p = qt , a = bt .
λ p + µ a λ ( qt ) + µ ( bt ) t ( λ q + µ b ) p a
Therefore, = = =t ≡ ≡ □
λ q + µb λ q + µb ( λ q + µb ) q b

16. 16

17. 17
B
§4: Basic Triangle Formulae
Cosine Rule
c h a
A C
b
x y
Figure 4.1
x y
Notice that: cos A = , cos C =
c a
Therefore, x = c cos A, y = a cos C
Hence b = x + y ≡ c cos A + a cos C
Using symmetry we interchange the variables to yield the complete set of results
thus: a
a = b cos C + c cos B
b = c cos A + a cos C (4.1)
c b
c = a cos B + b cos A
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:
a 2 = ab cos C + ac cos B (4.2)
b2 = bc cos A + ab cos C (4.3)
c 2 = ac cos B + bc cos A (4.4)
Now construct (4.3) + (4.4) - (4.2) to give:
b2 + c 2 − a 2 = 2bc cos A
Therefore,
a 2 = b2 + c 2 − 2bc cos A (4.5)
Equation (4.5) is known as the Cosine Rule for triangles.
Symmetry yields the other forms:
b2 = c 2 + a 2 − 2ca cos B and c 2 = a 2 + b2 − 2ab cos C

18. 18
Sine Rule
Also from Figure 4.1 we have the further set of relations:
h h
sin A = , sin C =
c a
Therefore, h = c sin A or h = a sin C
Hence, c sin A = a sin C ≡ h
sin A sin C
Therefore, =
a c
However, the initial orientation of the triangle ABC was arbitrary
sin A sin B sin C
⇒ = = (4.6)
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 Circumcircle of the triangle ABC)
B B
P
A
C
O O
A C P
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).
AC b
Therefore, from Figure 4.2 we have: sin ( APC ) ≡ sin ( B ) = ≡
AP 2 R
where R is the radius of the Circumcircle of triangle ABC.

19. 19
From Figure 4.3 APC = 180 − B (Cyclic Quadrilateral)
Therefore, sin ( APC ) = sin (180 − B ) ≡ sin B
Hence, as for Figure 4.2 we have
b sin B 1
sin B = or =
2R b 2R
Therefore, from (4.6) we have:
sin A sin B sin C 1
= = = (4.7)
a b c 2R
Tangent Rule
b sin B
From (4.7) we have that =
c sin C
Using Theorem (3.1) with λ ≡ m ≡ n = 1 and µ = −1 this relation can be written
(1) b + ( −1) c (1) sin B + ( −1) sin C
=
(1) b + (1) c (1) sin B + (1) sin C
b − c sin B − sin C
That is =
b + c sin B + sin C
Using (2.8) and (2.7) this can be rewritten as
B +C   B −C 
2 cos 
  sin  
b−c  2   2 ,
=
b + c 2 sin  B + C  cos  B − C 
   
 2   2 
B −C 
sin 
 
b−c  2 × 1
= ,
b + c cos  B − C  sin  B + C 
   
 2   2 
B+C 
cos 
 
 2 
B−C 
tan 
 
b−c  2 .
=
b + c tan  B + C 
 
 2 

20. 20
b−c
That is tan 1 ( B − C ) = tan 1 ( B + C ) (4.8)
b+c
2 2
However, A + B + C = 180 ⇒ 1
2 ( B + C ) = 90 − 1 A
2 (4.9)
From (4.9) we see that
tan 1 ( B + C ) = tan ( 90 − 1 A )
2 2
tan(90 ) − tan 1 A
= 2
1 + tan(90 ) tan 1 A
2
tan 2 A
1
1−
tan(90 )
=
1
+ tan 1 A
2
tan(90 )
1− 0
=
0 + tan 1 A
2
∴ tan 1 ( B + C ) = cot 1 A
2 2
Hence (4.8) can be written in its alternative form
b−c
tan 1 ( B − C ) = cot 1 A (4.10)
b+c
2 2
Equation (4.10) is known as the Tangent Rule for triangles.
Symmetry yields the other forms:
c−a
a
tan 1 ( C − A ) = cot 1 B
c+a
2 2
c b a−b
tan 1 ( A − B ) = cot 1 C
a+b
2 2

21. 21
§5: Other Triangle Formulae
B
Area Formulae in terms of a, b and c.
c h a
A C
b
Figure 5.1
The area of triangle ABC is found from
h
∆ = 1 bh, where sin C =
2 ,
a
= 1 b ( a sin C )
2
Hence, ∆ = 1 ab sin C
2 (5.1)
a
By symmetry, ∆ = 1 ab sin C = 1 bc sin A = 1 ca sin B
2 2 2
c b
c
From (4.7) we have that sin C =
2R
c 
Therefore, (5.1) becomes ∆ = 1 ab 
2  
 2R 
abc
That is: ∆= (5.2)
4R
abc
or R= .
4∆
a sin B
Also, from (4.7) we have that b=
sin A
Therefore, (5.1) now becomes:
a sin B 
∆ = 1 a
2   sin C
 sin A 
sin B sin C
That is: ∆ = 1 a2
2
sin A
Hence,
a
sin B sin C 1 2 sin C sin A 1 2 sin A sin B
∆ = 1 a2
2 = 2b = 2c (5.3)
c b
sin A sin B sin C

22. 22
Sin A in terms of a, b and c, etc.
From (1.2)
sin 2 A = 1 − cos 2 A
= (1) − ( cos A )
2 2
= (1 − cos A )(1 + cos A )
Therefore, from (4.5) this becomes
 b2 + c 2 − a 2   b2 + c 2 − a 2 
sin 2 A =  1 −  1 + 
 2bc  2bc 
  
 2bc − b 2 + c 2 − a 2   2bc + b 2 + c 2 − a 2

=  
 2bc  2bc 
  
 2bc − b 2 − c 2 + a 2   2bc + b 2 + c 2 − a 2

=  
 2bc  2bc 
 a − b + 2bc − c   b + 2bc + c − a 
2 2 2 2 2 2
=  
 2bc  2bc 
 a − (b − c )   (b + c ) − a 
2 2 2 2
=  
 2bc  2bc 
  
1
( )(
= 2 2 a2 − (b − c ) (b + c ) − a2
4b c
2 2
)
1
= 2 2 ( a − b + c )( a + b − c )( b + c − a )( b + c + a )
4b c
1
= 2 2 ( a + b + c − 2b )( a + b + c − 2c )( b + c + a − 2a )( b + c + a )
4b c
Now let s = 1 ( a + b + c ) , the Semi-perimeter, then we have
2
1
sin 2 A = ( 2s − 2b )( 2 s − 2c )( 2s − 2a )( 2s )
4b 2 c 2
1
= 2 2 16 ( s − b )( s − c )( s − a )( s )
4b c
Hence
4
sin 2 A = s ( s − a )( s − b )( s − c )
bc
2 2
Therefore,
2
a
sin A = s ( s − a )( s − b )( s − c )
bc
2
sin B = s ( s − a )( s − b )( s − c ) (5.4)
c b ca
2
sin C = s ( s − a )( s − b )( s − c )
ab

23. 23
Notice further that these three identities from (5.4) could be written
2a
sin A = s ( s − a )( s − b )( s − c ) ≡ 2aK
abc
2b
sin B = s ( s − a )( s − b )( s − c ) ≡ 2bK (5.5)
abc
2c
sin C = s ( s − a )( s − b )( s − c ) ≡ 2cK
abc
From which, the Sine Rule can be deduced, since
sin A sin B sin C
≡ 2K  ≡
1 
= =  
a b c  2R 
Moreover, from (5.1) and (5.4) we have that
2
∆ = 1 ab × sin C ≡ 1 ab ×
2 2 s ( s − a )( s − b )( s − c )
ab
Hence
∆ = s ( s − a )( s − b )( s − c ) (5.6)
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.
a
2∆ 2∆ 2∆
sin A = , sin B = , sin C = (5.7)
c b bc ca ab
Cos A, cos B, cos C in terms of a, b and c.
From (4.5) we simply rearrange to yield
a
b2 + c2 − a 2 c2 + a 2 − b2 a 2 + b2 − c2
cos A = , cos B = , cos C = (5.8)
c b 2bc 2ca 2ab
Tan A, tan B, tan C in terms of a, b and c.
From (1.1), (5.5) and (5.8) we have
a
4∆ 4∆ 4∆
tan A = , tan B = 2 , tan C = 2 (5.9)
c b b +c −a
2 2 2
c +a −b
2 2
a + b2 − c2

24. 24

25. 25
§6: Associated Circles
A
Incircle
Let I be the Incentre of the F
E
triangle ABC, obtained by
bisecting the interior angles of
the triangle ABC. Then I
ID = IE = IF ≡ r , where r is the
radius of the incircle.
( ID, IE and IF are the ⊥ 's . C
B D
from I to the respective sides)
Figure 6.1
We have that
Area of triangle ABC = Areas of triangles ( BIC + CIA + AIB )
∆ = 1 ar + 1 br + 1 cr
2 2 2
∆ = 1 r (a + b + c)
2
Hence for s, the semi perimeter
∆ = rs (6.1)
∆
or r= (6.2)
s
Now, let
AE = AF ≡ x  Tangents to a circle from

BD = BF ≡ y  a single point have equal

CD = CE = z  lengths
⇒ x+ y+ y+z+z+x= p ( perimeter )
Hence
x+ y+z =s ( semi perimeter )
But BD + DC = y + z ≡ a
⇒ x+a=s
Hence x = s−a.
Therefore, we can show: a
x=s−a
y = s−b (6.3)
c b
z = s−c

26. 26
A
From triangle ABC we see that
s-a s-a
IF r
tan ( 1 A ) = =
FA s − a
2
F
Similarly E
r
a
tan ( 1 A ) = s-b s-c
s−a
2
I
r
tan ( 1 B ) = (6.4)
s−b
2
c b
D C
r
tan ( 1 C ) = s-c
s−c B s-b
2
Figure 6.2
Moreover, from triangles AIE, BIF and CID we have
AE = AF ≡ s − a = r cot ( 1 A )
a
2
BD = BF ≡ s − b = r cot ( 1 B )
2 (6.5)
c b
CD = CE ≡ s − c = r cot ( 1 C )
2
Therefore, as a = BD + DC then
a = r cot ( 1 B ) + r cot ( 1 C )
2 2
= r ( cot ( 1 B ) + cot ( 1 C ) )
2 2
 cos ( 1 B ) cos ( 1 C ) 
= r
 sin ( 1 B ) + sin ( 1 C ) 
2 2

 2 2 
 cos ( 1 B ) sin ( 1 C ) + sin ( 1 B ) cos ( 1 C ) 
= r

2 2 2 2


 sin ( 1 B ) sin ( 1 C )
2 2 
sin ( 1 B + 1 C )
=r 2 2
sin ( 2 B ) sin ( 1 C )
1
2
But A + B + C = 180 and therefore, 1
2 B + 1 C = 90 − 1 A , giving
2 2
sin ( 90 − 1 A )
a=r
2
sin ( 1 B ) sin ( 1 C )
2 2
That is
cos ( 1 A )
a=r 2
.
sin ( 1 B ) sin ( 1 C )
2 2
Hence
sin ( 1 B ) sin ( 1 C )
r=a 2 2
(6.6)
cos ( 1 A )
2

27. 27
Together, we have
sin ( 1 B ) sin ( 1 C )
r=a 2 2
a cos ( 1 A )
2
sin ( 1 C ) sin ( 1 A )
r =b 2 2
(6.7)
c b cos ( 1 B )
2
sin ( 1 A ) sin ( 1 B )
r =c 2 2
cos ( 1 C )
2
From (4.7) we have
sin A 1
= , etc, where R is the Circumcircle.
a 2R
or
a = 2 R sin A , etc
Thus (6.6) becomes
sin ( 1 B ) sin ( 1 C )
r = 2 R sin A ⋅ 2 2
cos ( 1 A )
2
Using (2.11) gives
sin ( 1 B ) sin ( 1 C )
r = 2 R × 2sin ( A ) cos ( A ) ⋅
1 1 2 2
cos ( 1 A )
2 2
2
Hence
r = 4 R sin ( 1 A ) sin ( 1 B ) sin ( 1 C )
2 2 2 (6.8)

28. 28
Excircles
A
Let I A be the centre of a circle, opposite
angle A, of radius rA obtained by bisecting
the angles B and C externally and A
internally. I
We have that: B D A' D' C
Area of triangle ABC = E'
Areas of triangles ( AI A B + CI A A − BI AC )
That is
IA
∆ = 1 crA + 1 brA − 1 arA
2 2 2
F'
= 1 rA ( c + b − a )
2
= 1 rA ( a + b + c − 2a )
2
Figure 6.3
Hence for s, the semi perimeter
∆ = rA ( s − a ) (6.9)
Similarly, we have
∆
rA =
a s−a
∆
rB = (6.10)
c b
s−b
∆
rC =
s−c
Notice further that A, I and I A are collinear as both AI and AI A lie on the bisector
line of angle A
Now let
AE ' = AF ' ≡ x A  Tangents to a circle from

BD ' = BF ' ≡ y A  a single point have equal

CD ' = CE ' = z A  lengths
Then AF '+ AE ' = AB + BF '+ AC + CE '
2 xA = c + y A + b + z A
⇒
= b + c + ( yA + z A )
but y A + z A = BD '+ D ' C ≡ BC = a .

29. 29
Therefore:
2 xA = a + b + c
xA = 1 ( a + b + c )
2
Hence
xA = s ( semi perimeter ) (6.11)
A
In a similar manner
a
xB = s and xC = s
c b
The three distances x A = s, xB = s and xC = s
are displayed below D A'
B D' C
E'
A
IA
F'
IC
B C
Figure 6.4
Figure 6.5
IB
A
Figure 6.6
B C

30. 30
Notice also from triangle AI A F ' and from (6.11) that
I A F ' rA
tan ( 1 A ) =
2 =
AF ' s
⇒
rA = s tan ( 1 A ) .
2
Similarly
a
rA = s tan ( 1 A )
2
rB = s tan ( 1 B )
2 (6.12)
c b
rC = s tan ( 1 C )
2
Therefore, we can further show that
a AE ' = AF ' ≡ s = rA cot ( 1 A ) ,
2
s = rB cot ( 1 B ) ,
2 (6.13)
s = rC cot ( 1 C ) .
c b
2
Moreover, since BF ' = 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 incircle with side a, D ' is the point of
tangency of the excircle with side a, and where the symbol ∼ is the positive
difference between its two arguments. Namely:
DD ' = b ∼ c ≡ b − c (6.14)
Further, as BD ' = CD = s − c it follows that A ' is the midpoint of DD ' where A '
is also the midpoint of BC.

31. 31
Heron’s Area Formula
IA
C
D
I
D'
A
B
Figure 6.7
As BI and BI A bisect the angle B both internally and externally, it follows that
IBI A = 90 . Moreover, since IDB and I A D ' B are also right angles then this
implies that triangles BID and BI A D ' are similar.
Namely:
Triangles BID ≅ I A BD '
BI ID BD
Thus = =
I A B BD ' I A D '
BI r s −b
that is: = =
I AB s − c rA
Therefore: r ⋅ rA = ( s − b )( s − c ) (6.15)
On using (6.2) and (6.10) this becomes
∆ ∆
⋅ = ( s − b )( s − c )
s s−a
Hence ∆ = s ( s − a )( s − b )( s − c ) (6.16)
Which is the form of the area of triangle ABC as met previously in (MP.7) and
(5.6)

32. 32

33. 33
§7: Further Triangle Formulae
Y
IA
C
I
A D B X
Figure 7.1
Tan ( 1 A ) , tan ( 1 B ) and tan ( 1 C ) in terms of a , b and c.
2 2 2
ID r
From Figure 7.1 we have tan ( 1 A ) = ≡
AD s − a
2
On using (6.15) this becomes
 ( s − b )( s − c ) 
 
tan ( 1 A ) = 
rA  = ( s − b )( s − c )
s−a rA ( s − a )
2
On using (6.9) we have
( s − b )( s − c ) ( s − b )( s − c )
tan ( 1 A ) = =
2
∆ ∆
( s − a)
(s − a)
( s − b )( s − c ) ( s − b )( s − c )
Hence: tan ( 1 A ) = =
s ( s − a )( s − b )( s − c ) s (s − a)
2
Similarly,
( s − b )( s − c )
tan ( 1 A ) =
s(s − a)
2
a
( s − c )( s − a )
tan ( 1 B ) = (7.1)
s ( s − b)
2
c b
( s − a )( s − b )
tan ( 1 C ) =
s (s − c)
2

34. 34
Cos ( 1 A ) , cos ( 1 B ) and cos ( 1 C ) in terms of a , b and c.
2 2 2
From Figure 7.1 we have I ACY = 1
2 (180 − C)
= 90 − 1 C
2
= 1 ( A + B)
2
Also, using the construction properties of the Incircle and Excircle
I A AC = 1 A
2
Hence
AI AC = 180 − ( 1 A + (180 −
2 I ACY ) )
= 180 − ( 1 A + (180 − 1 ( A + B ) ) )
2 2
= 180 − (180 − 1 B )
2
AI AC = 1 B ≡
2 DBI
Therefore, triangles AIB and ACI A are similar, that is AIB ≅ ACI A .
AI IB AB
= =
AC CI A AI A
⇒
AI IB c
= =
b CI A AI A
Hence AI × AI A = bc (7.2)
AX
Also from Figure 7.1 cos ( 1 A ) =
2 (7.3)
AI A
AD
and cos ( 1 A ) =
2 (7.4)
AI
The product of (7.3) and (7.4) gives
AX AD
cos ( 1 A ) × cos ( 1 A ) =
2 2 ×
AI A AI
That is
AX × AD
cos2 ( 1 A ) =
AI A × AI
2
On using (6.3), (6.11) and (7.2) this becomes
s × (s − a)
cos 2 ( 1 A ) =
2
bc
Hence
s(s − a)
cos ( 1 A ) =
2
a bc
s ( s − b)
cos ( 1 B ) =
2 (7.5)
c b ca
s (s − c)
cos ( 1 C ) =
2
ab

35. 35
Sin ( 1 A ) , sin ( 1 B ) and sin ( 1 C ) in terms of a, b and c.
2 2 2
Using (1.1), (7.1) and (7.5) we have that
sin ( 1 A ) = tan ( 1 A ) cos ( 1 A )
2 2 2
Hence
( s − b )( s − c ) s(s − a)
sin ( 1 A ) = ×
s(s − a)
2
bc
Giving
( s − b )( s − c )
sin ( 1 A ) =
2
a bc
( s − c )( s − a )
sin ( 1 B ) =
2 (7.6)
c b ca
( s − a )( s − b )
sin ( 1 C ) =
2
ab
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.

36. 36

37. 37
§8: Further Triangle Relationships
Relationship between r and R A
Note that
BIC = 180 − 1 B − 1 C
2 2
I
= 180 − 1 ( B + C )
2 r
= 180 − 1 (180 − A )
2 B C
D
= 90 + A1
2 Figure 8.1
ID r
and that sin ( 1 B ) =
2 ≡
IB IB
Hence, r = IB sin ( 1 B )
2 (8.1)
Using the Sine Rule of (4.6) we have that
sin ( 1 C ) sin ( 90 + 1 A ) sin ( 1 B )
= =
2 2 2
,
IB BC IC
and from the first two of these fractions we have
sin ( 1 C ) cos ( 1 A )
2
= 2
.
IB BC
BC sin ( 1 C )
Therefore: IB = 2
,
cos ( 1 A )
2
a sin ( 1 C )
that is IB = 2
. (8.2)
cos ( 1 A )
2
Substituting (8.2) into (8.1) gives
 a sin ( 1 C ) 
r = cos ( 1 A )  sin ( 2 B )
2 1

 2 
a sin ( 2 C ) sin ( 1 B )
1
= 2
cos ( 2 A )
1
2 R sin A sin ( 1 B ) sin ( 1 C )
Using (4.7) a can be written as: a = 2 R sin A, giving r = 2 2
.
cos ( 1 A )
2
Using (2.11) gives
2 R × 2sin ( 1 A ) cos ( 1 A ) sin ( 1 B ) sin ( 1 C )
r= 2 2 2 2
.
cos ( 1 A )
2
Hence: r = 4 R sin ( 1 A ) sin ( 1 B ) sin ( 1 C )
2 2 2 (8.3)
as met previously in (6.8).

38. 38
Relationships between rA , rB , rC and R.
From Figure 8.2 we see that
C
I AD ' r
sin ( I A BD ') = ≡ A
I AB I AB D'
IA
rA
that is rA = I A B sin ( 90 − 1 B )
2
(8.4) rA = I A B cos ( 1 B )
2
A B
Figure 8.2
Using the Sine Rule of (4.6) we have that
sin ( 90 − 1 C ) sin ( 90 − 1 A ) sin ( 90 − 1 B )
= =
2 2 2
I AB BC I AC
and from the first two of these fractions we have
cos ( 1 C ) cos ( 1 A )
2
= 2
I AB BC
BC cos ( 1 C )
Therefore, I AB = 2
.
cos ( 1 A )
2
a cos ( 1 C )
that is, I AB = 2
(8.5)
cos ( 1 A )
2
Substituting (8.5) into (8.4) gives
 a cos ( 1 C ) 
 cos ( 1 A )  cos ( 2 B )
rA =  2 1

 2 
a cos ( 1 C ) cos ( 1 B )
= 2 2
cos ( 2 A )
1
Using (4.7) a can be written as: a = 2 R sin A ,
2 R sin A cos ( 1 C ) cos ( 1 B )
giving rA = 2 2
.
cos ( 1 A )
2
Using (2.11) gives
2 R × 2sin ( 1 A ) cos ( 1 A ) cos ( 1 C ) cos ( 1 B )
rA = 2 2 2 2
.
cos ( 1 A )
2
Hence: a rA = 4 R sin ( 1 A ) cos ( 1 B ) cos ( 1 C ) ,
2 2 2
and similarly rB = 4 R cos ( 1 A ) sin ( 1 B ) cos ( 1 C ) ,
2 2 2 (8.6)
c b
rC = 4 R cos ( 1 A ) cos ( 1 B ) sin ( 1 C ) .
2 2 2

39. 39
The distances AI , BI and CI .
From Figure 8.3 notice that C
ID IA
sin ( 1 A ) =
2 I
IA
A D B X
Therefore, Figure 8.3
a
AI = r cosec ( 1 A )
2
BI = r cosec ( 1 B )
2 (8.7)
c b
CI = r cosec ( 1 C )
2
If we just consider the triangle AI A X for the present and determine some of its
angles from Figure 8.4 IA
we find that:
Clearly AIB = 180 − 1 ( A + B )
2
I
and so on using the Sine Rule of (4.6)
we have from triangle AIB
A D B X
sin ( B ) sin ( AIB ) sin ( A )
1 1
2
= = 2 Figure 8.4
AI AB BI
Considering just the first two fractions here gives
AB sin ( 1 B )
AI = 2
sin ( AIB )
Therefore
c sin ( 1 B )
AI = 2
sin (180 − 1 ( A + B ) )
2
That is on using (2.2):
c sin ( 1 B )
AI = 2
sin ( 1 ( A + B ) )
2
c sin ( 1 B )
= 2
sin ( 90 − 1 C )
2
c sin ( 1 B )
= 2
cos ( 1 C )
2
But on using (4.7) c can be written as: c = 2 R sin C , giving

40. 40
2 R sin C sin ( 1 B )
AI = 2
cos ( 1 C )
2
2 R × 2sin ( 1 C ) cos ( 1 C ) sin ( 1 B )
Using (2.11) gives AI = 2 2 2
cos ( 1 C )
2
Hence
AI = 4 R sin ( 1 B ) sin ( 1 C )
a
2 2
BI = 4 R sin ( 1 C ) sin ( 1 A )
2 2 (8.8)
c b
CI = 4 R sin ( 1 A ) sin ( 1 B )
2 2
which give an alternative form of (8.7). Further, from (8.7), as it is somewhat
r
easier we may proceed AI = r cosec ( 1 A ) =
sin ( 1 A )
2
2
Thus, from (7.6) we have
r r bc
AI = =
( s − b )( s − c ) ( s − b )( s − c )
bc
r s ( s − a ) bc
≡
s ( s − a )( s − b )( s − c )
r s ( s − a ) bc s ( s − a ) bc
Using (5.6) and (6.1) AI = ≡ .
∆ s
( s − a ) bc (s − a)
Hence, AI = = bc
a s s
a ( s − b) c ( s − b)
similarly BI = = ca (8.9)
c b s s
ab ( s − c ) (s − c)
CI = = ab
s s
Note that the relations of (8.9) can be written
abc ( s − a ) (s − a)
AI = ≡K ,
s a a
abc ( s − b ) ( s − b) abc
BI = ≡K , where K = (8.10)
s b b s
abc ( s − c ) (s − c)
CI = ≡K .
s c c
(s − a) ( s − b) (s − c)
⇒ AI : BI : CI = : : (8.11)
a b c

41. 41
The distances AI A , BI B and CI C .
C
IA
I
A D B X
Figure 8.5
From Figure 8.5 notice that
I AD
sin ( 1 A ) =
2
IAA
Therefore,
AI A = rA cosec ( 1 A )
a
2
BI B = rB cosec ( 1 B )
2 (8.12)
c b
CI C = rC cosec ( 1 C )
2
C
Notice also from Figure 8.6 that
IA
θ + θ + φ + φ = 180
θ + φ = 90 I
Hence
IBI A = 90
Also, since
D B
θ = 1 B and φ = 1 (180 − B ) = 90 − 1 B
2 2 2
Figure 8.6
then I I A B = 90 − 1 ( A + B )
2
IA
I
Figure 8.7
A D B

42. 42
On using the Sine Rule of (4.6) we have from triangle ABI A
sin ( 90 + 1 B ) sin ( AI A B ) sin ( 1 A )
= =
2 2
AI A AB BI A
Considering just the first two fractions here gives
AB sin ( 90 + 1 B )
AI A =
2
sin ( AI A B )
Therefore
c cos ( 1 B )
AI A = 2
sin ( 90 − 1 ( A + B ) )
2
That is on using (2.2):
c cos ( 1 B )
AI A = 2
cos ( 1 ( A + B ) )
2
c cos ( 1 B )
= 2
cos ( 90 − 1 C )
2
c cos ( 1 B )
= 2
sin ( 1 C )
2
But on using (4.7) c can be written as: c = 2 R sin C
giving
2 R sin C cos ( 1 B )
AI A = 2
sin ( 2 C )
1
Using (2.11) gives
2 R × 2sin ( 1 C ) cos ( 1 C ) cos ( 1 B )
AI A = 2 2 2
sin ( 1 C )
2
Hence a AI A = 4 R cos ( 1 B ) cos ( 1 C )
2 2
and similarly BI B = 4 R cos ( 1 C ) cos ( 1 A )
2 2 (8.13)
c b
CI C = 4 R cos ( 1 A ) cos ( 1 B )
2 2
which give an alternative form of (8.12)
Further, from (8.12), as it is slightly easier we may proceed
rA
AI A = rA cosec ( 1 A ) =
sin ( 1 A )
2
2

43. 43
Thus, from (6.10) and (7.6) we have
∆ 1
AI A =
( s − a ) ( s − b )( s − c )
bc
s ( 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:
bc
or AI A =
(s − a)
s
a
ca
similarly BI B = (8.14)
c b ( s − b)
s
ab
CI C =
(s − c)
s
In a manner similar to the processes used to arrive at (8.11) we can also show that
1 1 1
AI A : BI B : CI C = : : (8.15)
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:
a
AI × AI A = bc
BI × BI B = ca (8.16)
c b
CI × CI C = ab

44. 44
The distances II A , II B and II C . IA
I
Y
A D B X
Figure 8.8
Consider the similar triangles AID and II AY , we have
AI AD ID
= =
II A DX I AY
That is
AI (s − a) r
= =
II A AX − AD I A X − YX
AI (s − a) r
= =
II A s − ( s − a ) rA − ID
AI ( s − a ) r
Hence, = = (8.17)
II A a rA − r
From the first two fractions we have
a 
II A = 
  AI
a
s−a
b 
similarly II B = 
  BI (8.18)
c b  s−b
c 
II C = 
  CI
s−c
Therefore, on using the values from the first forms in (8.10), the results in (8.18)
can be written in the manner
a  abc ( s − a )
II A = 
 ×
s−a s a
abc a
II A =
s (s − a)
Hence
abc a abc b abc c
II A = , II B = , II C = . (8.19)
s (s − a) s ( s − b) s (s − c)

45. 45
Squaring the value of II A from (8.19) gives
 abc   a 
( II A )
2
=  
 s  (s − a) 
Therefore, from the last two fractions of (8.17) this becomes
abc  rA − r 
( II A ) =
2
   (8.20)
 s  r 
abc
From (5.2) we have ∆=
4R
and from (6.1) that ∆ = rs
abc
Hence, 4 Rr = (8.21)
s
Therefore, from (8.20) and (8.21) we have that
( II A ) = 4 R ( rA − r ) , ( II B ) = 4 R ( rB − r ) , ( IIC ) = 4 R ( rC − r )
2 2 2
(8.22)
Moreover, using the half-angle formula in (7.5) we see that (8.19) can be
rearranged thus:
a 2bc bc 1 a
II A = =a =a ≡
s(s − a) s(s − a) s ( s − a ) cos ( 1 A )
2
bc
Hence
a b c
II A = , II B = , II C = . (8.23)
cos ( A )
1
2 cos ( B )
1
2 cos ( 1 C )
2
Therefore, on using the results of (4.7), (8.23) can be rewritten in a further,
alternative form
II A = 4 R sin ( 1 A ) ,
2 II B = 4 R sin ( 1 B ) ,
2 II C = 4 R sin ( 1 C ) .
2 (8.24)

46. 46

47. 47
§9: Further Triangle Centres
The Orthocentre of any Triangle ABC
The perpendiculars drawn from the vertices of a triangle ABC to the opposite
sides are concurrent at a point called the Orthocentre, H. A
From Figure 9.1 let AD, BE and CF be the perpendiculars
F
on BC, CA and AB respectively, and H the Orthocentre;
then
DH = BD tan ( HBD )
E
H
= AB cos B × tan ( 90 − C ) ,
= c cos B cot C
Using (4.7) c can be written as: B D C
Figure 9.1
c = 2 R sin C ,
giving DH = 2 R sin C cos B cot C
Hence a
DH = 2 R cos B cos C hA
EH = 2 R cos C cos A hB (9.1)
c b
FH = 2 R cos A cos B hC
DA = AB sin B
Further, = c sin B
= 2 R sin C sin B
That is a
DA = 2 R sin B sin C
EB = 2 R sin C sin A (9.2)
c b
FC = 2 R sin A sin B
Therefore, from (9.2) and (9.1) we have that
HA = DA − DH
= 2 R sin B sin C − 2 R cos B cos C
Using (2.3) this becomes
HA = −2 R cos ( B + C )
= −2 R cos (180 − A )
Hence:
a
HA = 2 R cos A
HB = 2 R cos B (9.3)
c b
HC = 2 R cos C

48. 48
Notice that
HA2 = ( 2 R cos A )
2
= 4 R 2 cos 2 A
= 4 R 2 (1 − sin 2 A )
= 4 R 2 − 4 R 2 sin 2 A
= 4 R 2 − ( 2 R sin A )
2
Using (4.7) a can be written as: a = 2 R sin A ,
giving HA2 = 4 R 2 − a 2
Hence, we have the alternative forms of (9.3) as
a
HA = 4 R 2 − a 2
HB = 4 R 2 − b2 (9.4)
c b
HC = 4 R 2 − c 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 hA , hB and hC .
Area of triangle ABC = 1 BC × AD
2
= 1 a × hA
2
a
Therefore,
2∆ 2∆ 2∆
hA = , and similarly hB = , hC = .
a b c c b