TITLE: Another beauty hidden in a simple triangle
QUESTION [6 upvotes]: Inscribed circle of triangle $ABC$ touches sides at points $D,E,F$. Reflect points $D,E,F$ with respect to angle bisectors $Ax,By,Cz$ and denote new points with $D',E',F'$. Denote midpoints of triangle sides with $D'',E'',F''$.
Prove that lines $D'D'',E'E'',F'F''$ meet at point $G$ that belongs to the inscribed circle.
It's relatively easy to prove that $D'D'',E'E'',F'F''$ really concur and I can demonstrate it if necessary. However the fact that the point of intersection $G$ also lies on the inscribed circle is something that makes this problem even more interesting. And harder - I don't have that part of the solution yet.

REPLY [4 votes]: You may consider that $D',E',F'$ lie on the incircle while $D'',E'',F''$ lie on the nine-point circle of $ABC$. These circles are tangent by Feuerbach's theorem, hence if you have already proved that $D'D'',E'E'',F'F''$ are concurrent, a natural candidate for their commont point is the perspector of the incircle and the nine-point circle, i.e. the Feuerbach point of $ABC$. I leave to you to prove this claim by exploiting trilinear/barycentric coordinates (which is a lengthy approach, but a no-brainer) or more subtle arguments (like the fact that the sides of $D'E'F'$ and the sides of $D''E''F''$ are parallel :) )
Nice problem, by the way, since it provides a straightforward construction of the Feuerbach point of a triangle (alternative to intersecting the incircle with the ray joining $I$ with the midpoint of $OH$). Is it taken from/inspired by a question from a past competition?
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