In a triangle $\Delta ABC$, let $X,Y$ be the foot of perpendiculars drawn from $A$ to the internal angle bisectors of $B$ and $C$. Prove that $XY$ is parallel to $BC$.
It works for an equilateral triangle because the angular bisector is also the perpendicular bisector.
I tried drawing a diagram to get some idea,
To prove that $XY$ is parallel to $BC$, i need to show that $\angle AFG=\angle AXY$ and $\angle AYF=\angle AGF$