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.