In triangle $ABC,$ let the angle bisectors be $\overline{BY}$ and $\overline{CZ}$. Given $AB = 14$, $AY = 14$, and $CY = 8$, find $BZ$.
A
14-x 14
Z Y
x 8
B C
We already found that BC = 8
Since CZ is a bisector
BZ / BC = AZ / AC
x / 8 = (14 -x) / 22
22x = 8 (14 -x)
22x = 112 - 8x
30 x = 112
x = 112/30 = 56 / 15 = BZ
+868 
