103 Trigonometry Problems

30 103 Trigonometry Problems
A
F
P
E
B
D1
Figure 1.29.
C
Assume that part (3) is true. Let segments BE and CF meet at P , and let ray AP
meet segment BC at D 1 (Figure 1.29). It suffices to show that D = D 1 . Cevians
AD 1 ,BE, and CF are concurrent at P . By our discussions above, we have
|AF |
|FB| · |BD 1|
|D 1 C| · |CE|
|EA| = 1 = |AF |
|FB| · |BD|
|DC| · |CE|
|EA| ,
implying that |BD 1|
|D 1 C| = |BD|
|DC| . Because both D and D 1 lie on segment BC, we conclude
that D = D 1 , establishing part (1).
Using Ceva’s theorem, we can see that the medians, altitudes, and angle bisectors
of a triangle are concurrent. The names of these concurrent points are the centroid
(G), orthocenter (H ), and incenter (I), respectively (Figure 1.30). If the incircle
of triangle ABC touches sides AB, BC, and CA at F , D, and E, then by equal
tangents, we have |AE| =|AF |, |BD| =|BF|, and |CD| =|CE|. ByCeva’s
theorem, it follows that lines AD, BE, and CF are concurrent, and the point of
concurrency is called the Gergonne point (Ge) of the triangle. All these four points
are shown in Figure 1.30. Given an angle, it is not difficult to see that the points
lying on the bisector of the angle are equidistant from the rays forming the angle.
Thus, the intersection of the three angle bisectors is equidistant from the three sides.
Hence, this intersection point is the center of the unique circle that is inscribed in
the triangle. That is why this point is the incenter of the triangle.
A
A
F
I
Ge
E
G
H
B
D
C
Figure 1.30.
B
C
Note that Ceva’s theorem can be generalized in a such a way that the point of
concurrency does not necessarily have to be inside the triangle; that is, the cevian