+0

# Angle bisectors

0
6
1
+1279

In triangle $ABC,$ let the angle bisectors be $\overline{BY}$ and $\overline{CZ}$. Given $AB = 12$, $AY = 12$, and $CY = 6$, find $BZ$.

Dec 18, 2023

#1
+222
+1

By the Power of a Point, [[C, YZ] = [A, YZ] + [B, YZ].]

Since quadrilateral BCZY is cyclic (as any angle bisector in a triangle divides an opposite side into segments proportional to the other two sides), we have ∠CYZ=∠CBZ, so [C,YZ]=[B,YZ].

Therefore, [[A, YZ] = [C, YZ] = [B, YZ].]

Furthermore, the circumcircle of triangle ABY has radius AY=12, so [A,YZ]=2AB⋅BY​=212⋅12​=72.

By the Angle Bisector Theorem, [\frac{BZ}{CZ} = \frac{AB}{AC} = \frac{12}{18} = \frac{2}{3}.]

Then [\frac{BZ}{BZ + CZ} = \frac{2}{5}.]

By the Power of a Point, and using that triangles BYZ and CYZ are similar, we can comput

[[B, YZ] = \frac{BZ \cdot BY}{2} = \frac{BZ \cdot \frac{2}{3} \cdot CZ}{2} = \frac{BZ \cdot CZ}{3}.]

Therefore, [\frac{BZ \cdot CZ}{3} = \frac{2}{5},]so BZ⋅CZ=56​.

Finally, by Stewart's Theorem,

[BC^2 = BZ^2 + CZ^2 - 2 \cdot BZ \cdot CZ \cdot \cos \angle BZC = BZ^2 + CZ^2 - \frac{12}{5}.]

Therefore,

[BC = \sqrt{BZ^2 + CZ^2 - \frac{12}{5}} = \sqrt{\frac{6}{5} + \frac{6}{5} - \frac{12}{5}} = \boxed{6}.]

Dec 19, 2023