From Lemmas in Olympiad Geometry, Titu Andreescu, Epsilon 8.6:
Let $ABC$ be a triangle and let $D,E,F$ be the incircle tangency points. Let the incircle intersect $AI, BI, CI$ at $M,N,P$ respectively.Prove that the Simson's Line of any point on incircle with respect to $DEF$ and $MNP$ are perpendicular.
My Progress:
We know that $M$ is midpoint of arc $EF$ and similarly. If $G$ is any point on the circle with respect to which simson lines are drawn, then $KG||AI, GJ||CI, GH||BI$. I'm thinking there could be a possible homothety because of parallel lines.
I also tried a Steiner line claim, by invoking orthocentres of triangle $DEF, MNP$. However I haven't proceeded much.
Thanks in advance
Since $E$ is the midpoint of $E'E''$ and $\angle A=90^\circ$, $\triangle AEE'$ is isosceles at $E$. It follows that $\angle AEE''=2 \angle EAE'$, equivalently $E$ trisects the small arc $AC$.
Best Answer
First, prove the following lemma: Let $ABC$ be inscribed in a circle $\omega$. Let $P$ be a point on $\omega$ and let $Q$ be a point on $\omega$ such that $PQ \perp BC$. Then $AQ$ is parallel to Simson line of $P$ with respect to $ABC$.
Next, draw chords $GX$ and $GY$ perpendicular to $EF$ and $MN$, respectively. By lemma, you have to prove that $DX \perp PY$. This can be shown by angle chasing.