Problem - 3934
In $\triangle{ABC}$, let $\angle{A}=120^\circ$. If $A'$, $B'$ and $C'$ are feet of the three interior angle bisectors as shown, prove $A'B'\perp A'C'$.
Extend $BA$ and consider $\triangle{AA'B}$. It is easy to see that $AC$ bisects $\angle{A}$'s exterior angle. Given $BB'$ bisects $\angle{B}$, it must hold that $A'B'$ bisects $\angle{CA'A}$ which is the exterior angle of $\angle{AA'B}$.
By a similar reasoning, we find $A'C'$ bisects $\angle{AA'B}$. Hence $$\angle{B'A'C'}=\frac{1}{2}\cdot \angle{CA'B}=90^\circ$$