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]
+289 https://artofproblemsolving.com/texer/xjsiitim
Question properly displayed here. Click "With bbcode"
+2 You can update the release date yourself: log in to your GEO account, go to your Series (GSExxxx) record, and use the 'UPDATE' button which is located at the top of the page; on the 'Data Field' page, there is a 'Data Release Date' box you can use to specify the new release date.
+2668 
By the Angle Bisector Theorem, we have AYAB=YCCB. Plugging in the given info, we have 68=3CB, meaning CB=4
Now, apply the Angle Bisector Theorem again: AZAC=ZBCB. Subbing in what we know, we have AB9=ZB4 (i).
For simplicity, let AB=x and ZB=y. We also know that x+y=8 (ii).
Solving this system, we find x=AB=7213 and y=ZB=3213
