+1471 In triangle $ABC,$ let the angle bisectors be $\overline{BY}$ and $\overline{CZ}$. Given $AB = 14$, $AY = 14$, and $CY = 8$, find $BC$.
+1768 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.
Let CZ=y. Then CY=8y. Since CY+YZ=14, y+8y=214=7. Therefore, [BC = CY + BZ = 8y + 2 = 8 \cdot 7 + 2 = \boxed{13.0909}.]
