103 Trigonometry Problems

1. Trigonometric Fundamentals 31
can be considered as a segment joining a vertex and a point lying on the line of the
opposite side. The reader might want to establish the theorem for the configuration
shown in Figure 1.31.
A
F
E
P
B
Figure 1.31.
C
D
With this general form in mind, it is straightforward to see that in a triangle, the
two exterior angle bisectors at two of its vertices and the interior angle bisector at
the third vertex are concurrent, and the point of concurrency is the excenter of the
triangle opposite the third vertex. Figure 1.32 shows the excenter I A of triangle ABC
opposite A. Following the reasoning of the definition of the incenter, it is not difficult
to see that I A is the center of the unique circle outside of triangle ABC that is tangent
to rays AB and AC and side BC.
C
IA
A
Figure 1.32.
B
The following example is another good application of Ceva’s theorem.
Example 1.12. [IMO 2001 Short List] Let A 1 be the center of the square inscribed in
acute triangle ABC with two vertices of the square on side BC (Figure 1.33). Thus
one of the two remaining vertices of the square lies on side AB and the other on
segment AC. Points B 1 and C 1 are defined in a similar way for inscribed squares with
two vertices on sides AC and AB, respectively. Prove that lines AA 1 ,BB 1 ,CC 1
are concurrent.