Solution to problem 4975 in Romantics of Geometry

This problem was proposed by Tran Viet Hung (Vietnam). Here is my proof. 

Proof for a). 

Perform an inversion around circle, $w$, centred at $A$ with radius $AE=AD$. The incircle is fixed since it is orthogonal with $w$. Touch points $E$ and $D$ are also fixed since they are on the circumference of $w$. The inverted image of $I$ is the intersection of $AI$ with $ED$, denoted $I'$. $B$ and $C$ are transformed into $B'$ and $C'$ on lines $AB$ and $AC$, respectively. The circumcircle $(ABC)$ is sent to the line passing through $B'$, $C'$. Line $BC$ is sent to the circle $(AB'C')$ and is tangent to the incircle at $P'$, the inverted image of $P$. The point $M$ is sent to $M'$, the intersection of $AI$ and $B'C'$. The line $MP$ is sent to the circle passing through $A$, $M'$ and $P'$. Finally, the inverted image of $Q$ is the second intersection of line $B'M'$ with $(AM'P')$ (See the inverted diagram above).

As angles are preserved under inversion, we want to show that $Q'$ lies on $E'D'$, taking advantage of the fact that $\angle{AI'E'}=90^\circ$. Let's supose that $E'D'$ cuts $B'C'$ in $Q''$. It suffices to show $AM'P'Q''$ is cyclic. Indeed, if $N$ is the intersection of $AI'$ with arc $B'C'$,  then, $N$ is the midpoint of arc $B'C'$, and it is known that $Q''$, $P'$ and $N$ are collinear (notice that the incircle turns out to be the $A$-mixtilinear excircle of $\triangle{AB'C'}$). A simple angle chase shows that $\angle{M'AP'}=\angle{M'Q''P'}$, implying $AM'P'Q''$ is cyclic and that $Q''=Q'$, as desired.

$\square$

Proof for b).

Under the same inversion we perfom in part a), the $A$-mixtilinear excircle is sent to the incircle of $\triangle{AB'C'}$ which touches $B'C'$ in $X''$, the inverted image of $X'$. The intersection of $AI$ with $BC$ is sent to $N$, the same midpoint of arc $B'C'$ (not containing $A$) we described in part a). It suffices to show that $Q'NX''A$ is cyclic. Indeed, we proved already that $Q'$ lies on $B'C'$, so if we prove that $\angle{X''Q'N} = \angle{X''AN}$ we are done. But it is well known that $AX''$, $AP'$ are isogonal and $Q'$, $P'$ and $N$ are collinear, so $\angle{X''AN} = \angle{M'AP'} = \angle{M'Q'P'} = \angle{X''Q'N}$, implying that $Q'NX''A$ is cyclic, as desired.

$\square$