Let $I$ be the incentre of triangle $ABC$. Let $D,E,F$ be the intersections between the incenter and sides $BC,CA,AB$ respectively. Let $M$ be the midpoint of $EF$ and let $Q$ be the second intersection between $AD$ and the incircle. Show that $MIDQ$ is a cyclic quadrilateral.
I have tried some angle-chasing, but no result. Any help?
Does this work? Let $\Gamma_{P,C(XYZ)}$ denote the power of $P$ wrt the circle through points $X,Y,Z$.
$AM \perp FE$ and $IM \perp FE$, thus $I,M,F$ are collinear. Thus $APDI$ is a cyclic quadrilateral $\Leftrightarrow |AM|\cdot |AI| = |AP| \cdot |AD|$. Now $|AP| \cdot |AD| = \Gamma_{A,C(IME)}=|AE|^2$ since $A$ is tangent to $C(IME)$ in $E$ ($IE$ is a diameter since $\angle IME=90^{\circ}$, and $IE \perp AE$).
Similarly $|AM|\cdot |AI|=\Gamma_{A,C(IMF)}=|AF|^2$. Clearly $|AF|=|AE|$ so we're done.
By looking at the green hi-lited figure, you should be able to see $\alpha = \beta$.
Construction: Extend QM to cut the circle at R. Join DR.
By looking at the blue hi-lited figure, you should be able to see $\beta = \gamma$, provided that $\angle DPI = 90^0$.
Result then follows. Hope the above can help.
