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For

Larousse Charlot


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.

Solution:

Consider the triangles DCAA and DAAB.  We know that BB is the bisector of the angle ÐB.  Similarly CC is the bisector of the angle ÐC.  If we let AB be the bisector of an exterior angle the triangle ACA, then C is the excenter of triangle ACA.  Hence, AC is the bisector of ÐAAB.

 

Similarly, let AC be the bisector of an exterior angle of triangle AAB.  Then, B is the excenter of triangle AAB, and AB is the bisector of angle ÐAAC.

Take notice that the sum of angles ÐCAA and ÐAAB is 180.  Also that ÐAAC= ÐCAB; similarly,

ÐCAB = ÐBAA.

So, we then have the following

180 = ÐCAB + ÐBAA  + ÐAAC + ÐCAB

      = 2ÐBAA  + 2ÐAAC

Multiply the equation by ½ and we obtain that

90 = ÐBAA  + ÐAAC

which is right.  Thus, the triangle DBAC is a right triangle.

Click here to see the triangle


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