I is the incenter of triangle $ABC$. $X$ and $Y$ are the feet of the perpendiculars from $A$ to $BI$ and $CI$. Prove that $XY$ is parallel to $BC$
I tried to use the angles $AXI$ and $AYI$ to prove them equal to angle $B/2$ and $C/2$ but I could not do that because I am unable the $AXI$ and $AYI$ accordingly to prove that they are alternate equal to $B/2$ and $C/2$.
Let the lines $AX,AY$ meet $BC$ at $U,V$ respectively.
Since $CY$ bisects $\angle C$ and is perpendicular to $AV$, $Y$ must be the midpoint of $AV$. Similarly, $X$ must be the midpoint of $AU$. So $XY$ is the midline of the triangle $AUV$ which means it is parallel to $UV$ and hence to $BC$.