THE PROBLEM: Let
angle A of triangle ABC have measure 120 degrees.
Let A' be the point of intersection of segment BC and the angle
bisector of angle A.
Likewise B' is the intersection of AC and the angle bisector of angle
B, and C' is the intersection of AB and the angle bisector of angle C. Prove angle B'A'C' is always a right angle.
Consider the triangles DCAAÕ and DAAÕB. We know that BBÕ is the bisector of the angle ÐB. Similarly CCÕ is the bisector of the angle ÐC. If we let AB be the bisector of an exterior angle the triangle ACAÕ, then CÕ is the excenter of triangle ACAÕ. Hence, AÕCÕ is the bisector of ÐAAÕB.
Similarly, let AC be the bisector of an exterior angle of triangle AAÕB. Then, BÕ is the excenter of triangle AAÕB, and AÕBÕ is the bisector of angle ÐAAÕC.
Take notice that the sum of angles ÐCAÕA and ÐAAÕB is 180. Also that ÐAAÕCÕ= ÐCÕAÕB; similarly,
ÐCAÕBÕ = ÐBÕAÕA.
So, we then have the following
180 = ÐCAÕBÕ + ÐBÕAÕA + ÐAAÕCÕ + ÐCÕAÕB
= 2ÐBÕAÕA + 2ÐAAÕCÕ
Multiply the equation by ½ and we obtain that
90 = ÐBÕAÕA + ÐAAÕCÕ
which is right. Thus, the triangle DBÕAÕCÕ is a right triangle.¬
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