Geometry with Trigonometry

142 Trigonometry; cosine and sine; addition formulae Ch. 9

so that

sin^2 α

a^2

=

4 b^2 c^2 −(b^2 +c^2 −a^2 )^2

4 a^2 b^2 c^2

=

2 (b^2 c^2 +c^2 a^2 +a^2 b^2 )−(a^4 +b^4 +c^4 )

4 a^2 b^2 c^2

As the right-hand side here is symmetrical ina,bandcwe must have

sin^2 α

a^2

=

sin^2 β

b^2

=

sin^2 γ

c^2

As the sines of wedge-angles are all positive, we may take square roots here and the
result follows.

9.5.3

In a triangle[A,B,C], let the mid-line of|BACmeet[B,C]at D and let d 1 =|A,D|.
Then

d 1 =

2 bc

b+c

cos

1

2

α.

Proof.By5.5

|B,D|

|D,C|

=

c

b

,

so that
|B,D|=

c

b+c

a.

On applying the sine rule to the triangle[A,B,D]we have that

d 1

sinβ

=

ca

b+c

1

sin^12 α

,

and so

d 1 =

ca

b+c

sinβ

sin^12 α

=

ca

b+c

sinβ

b

b

sin^12 α

=

ca

b+c

sinα

a

b

sin^12 α

=

2 bc

b+c

cos

1

2

α.

9.5.4 The Steiner-Lehmus theorem, 1842 ...............

Suppose that we are given a tri-
angle[A,B,C], that the mid-line
of|CBAmeetsCAatE, that the
mid-line of|ACBmeetsABatF,
and that|B,E|=|C,F|.Wethen
wish to show that the triangle is
isosceles. This is known as the
Steiner-Lehmus theorem.

A

B C

F E

Figure 9.13. Steiner-Lehmus theorem.