In a triangle $\Delta ABC$, let $X,Y$ be the foot of perpendiculars drawn from $A$ to the internal angle bisectors of $B$ and $C$

Question

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 $\angle AYF=\angle AGF$

Answer 1

Since $CX$ is a bisector and $AX\perp CX$, $X$ is the midpoint of $AF$. In a similar way we have that $Y$ is the midpoint of $AG$. By Thales' theorem, $XY\parallel FG$, hence $XY\parallel BC$.