103 Trigonometry Problems

20 103 Trigonometry Problems
A
A
B
P
D
C
B
C
D
P
D
D
C
P
P
C
A B A B
Figure 1.20.
In general, if P is a point on segment BC, then |AD| =|AP | sin ̸
|AP |·|BC| sin ̸ AP
B
AP B. Hence
[ABC] =
2
. More generally, let ABCD be a quadrilateral (not
necessarily convex), and let P be the intersection of diagonals AC and BD, as shown
|AC|·|BP| sin ̸ AP B
in Figure 1.20. Then [ABC] =
2
and [ADC] = |AC|·|DP| sin ̸ AP D
2
.
Because ̸ AP B + ̸ AP D = 180 ◦ , it follows that sin ̸ AP B = sin ̸ AP D and
|AC| sin ̸ AP
[ABCD] =[ABC]+[ADC] =
2
= |AC|·|BD| sin ̸ AP B
.
2
B
(|BP|+|DP|)
Now we introduce Ptolemy’s theorem: Inaconvexcyclic quadrilateral ABCD
(that is, the vertices of the quadrilateral lie on a circle, and this circle is called the
circumcircle of the quadrilateral),
|AC|·|BD|=|AB|·|CD|+|AD|·|BC|.
There are many proofs of this very important theorem. Our proof uses areas. The
product |AC|·|BD| is closely related to [ABCD]. Indeed,
[ABCD] = 1 2 ·|AC|·|BD| sin ̸ AP B,
where P is the intersection of diagonals AC and BD. (See Figure 1.21.)