In triangle $ABC,$ let the angle bisectors be $\overline{BY}$ and $\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BC$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$I$",I,SW); label("$Y$",Y,NW); label("$Z$",Z,S); [/asy]
In triangle $ABC,$ let the angle bisectors be $\overline{BY}$ and $\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BZ$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$I$",I,SW); label("$Y$",Y,NW); label("$Z$",Z,S); [/asy]
https://artofproblemsolving.com/texer/xjsiitim
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By the Angle Bisector Theorem, we have \({AY \over AB} = {YC \over CB}\). Plugging in the given info, we have \({6 \over 8} = {3 \over CB}\), meaning \(CB = 4\)
Now, apply the Angle Bisector Theorem again: \({AZ \over AC} = {ZB \over CB}\). Subbing in what we know, we have \({AB \over 9} = {ZB \over 4} \ \ \ (i)\).
For simplicity, let \(AB = x \) and \(ZB = y \). We also know that \(x + y = 8 \ \ \ (ii)\).
Solving this system, we find \(x = AB = {72 \over 13}\) and \(y = ZB = \color{brown}\boxed{32 \over 13}\)