In triangle $ABC,$ let the angle bisectors be $\overline{BY}$ and $\overline{CZ}$. Given $AB = 14$, $AY = 14$, and $CY = 8$, find $BZ$.
By the Angle Bisector Theorem, [ \frac{AZ}{ZB} = \frac{AC}{BC} = \frac{CY}{BY} = \frac{8}{14-Y}. ] Since AY+Y=14, [ \frac{AZ}{ZB} = \frac{8}{14-AY} = 8. ] Let ZB=x. Then AZ=8x. Since AZ+ZB=14, x+8x=14, so x=2.
Therefore, BZ=2.
Thanks!
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