Class Page

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 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.¬

Click here to see the triangle