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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,

enter image description here

To prove that $XY$ is parallel to $BC$, i need to show that $\angle AFG=\angle AXY$ and $\angle AYF=\angle AGF$

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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 $XY\parallel BC$.

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  • $\begingroup$ Can you show how "$X$ is the midpoint of $AF$"? $\endgroup$ – Aditya Dev Mar 2 '16 at 17:33
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    $\begingroup$ @AdityaDev: $CXA$ and $CXF$ are congruent by $SAA$, hence $AX=XF$. $\endgroup$ – Jack D'Aurizio Mar 2 '16 at 17:35
  • $\begingroup$ Which two sides are equal? The triangle is not equilateral. Although it looks like one. $\endgroup$ – Aditya Dev Mar 4 '16 at 4:00

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