103 Trigonometry Problems

1. Trigonometric Fundamentals 19
Note that this fact can also be obtained by extending ray OB to meet ω at D, and
then working on right triangle ABD.
A direct application of the law of sines is to prove the angle-bisector theorem:
Let ABC be a triangle (Figure 1.19), and let D be a point on segment BC such that
̸ BAD = ̸ CAD. Then
|AB|
|AC| = |BD|
|CD| .
Applying the law of sines to triangle ABD gives
|AB|
sin ̸ ADB = |BD|
sin ̸ BAD , or |AB|
|BD| = sin ̸ ADB
sin ̸ BAD .
|CD| = sin ̸ ADC
sin ̸
Similarly, applying the law of sines to triangle ACD gives |AC|
. Because
CAD
sin ̸ ADB = sin ̸ ADC and sin ̸ BAD = sin ̸ CAD, it follows that |AB|
|BD| = |AC|
|CD| ,
as desired.
This theorem can be extended to the situation in which AD 1 is the external bisector
of the triangle (see Figure 1.19).We leave it to the reader to state and prove this version
of the theorem.
A
A
B
D
C
B
Figure 1.19.
C
D1
Area and Ptolemy’s Theorem
Let ABC be a triangle, and let D be the foot of the perpendicular line segment from
A to line BC (Figure 1.20). Then [ABC] = |BC|·|AD|
2
. Note that |AD| =|AB| sin B.
Thus [ABC] =
|BC|·|AB| sin B
2
=
ac sin B
2
.