A geometry problem with the reflection of the incenter Consider $\triangle ABC$ and its incenter $I$. Let $M$ be the midpoint of arc $BC$ (not containing $A$) and $P$ be the second intersection point of the circumcircle of $\triangle ABC$ and the circle with center $I$ and radius $IA$. Prove that if $I'$ is the reflection of $I$ with respect to $BC$, then $I', P, M$ are collinear (lie on the same line). 
Please help, I am lost, I don't have any idea! Thanks a lot in advance! Anyway, this problem is based on a problem connected to the reflection of the incenter, which appeared in Mathematical Excalibur.
For help, here is a figure:

 A: Denote circumcircle of $\triangle ABC$ as $\Omega$ and circle with center $M$ and radius $MI$ as $\Gamma$. It's well-known that $\Gamma$ is the circumcircle of $\triangle BIC$.
Let $\Phi$ be the inversion with respect to the circle $\Gamma$. Then, for any point $X\neq M$ of the plane denote its image under the $\Phi$ as $X^*=\Phi(X)$.
Since $I$ and $J$ (here $J=I'$) are symmetric with respect to the line $BC$ their images $I^*$ and $J^*$ are symmetric with respect to $\Phi(BC)=\Omega$ (because inversion preserves symmetry eith respect to generalized circles; $\Phi(\{I,B,C\})=\{A,B,C\}$). However, $\Phi(I)=I$, so $J^*$ is the image of $I$ under the inversion with respect to the circle $\Omega$. In particular, $J^*$ lies on the line $OI$.
It's clear that $A$ and $P$ are symmetric with respect to $OI$. Hence, if $D$ is the point symmetric to $M$ with respect to $OI$, then $D\in\Omega$, $D, I, P$ are collinear and lines $MP$, $AD$ and $OI$ are concurrent. Denote the intersection of $MP$, $AD$ and $OI$ as $K$.
Now note that the collinearity of $P$, $M$ and $J$ is equivalent to the collinearity of $P$, $M$ and $J^*$ (since $M$ is the center of $\Gamma$). Thus, we need to prove that $J^*=K$.
Indeed, we have cyclic quadrilateral $ADMP$ with $AP\parallel MD$, where $K=AD\cap PM$, $I=AM\cap PD$. It means that $I$ and $K$ are inverse images of each other with respect to $\Omega$. Therefore, $J^*=K$, as desired.

A: @Alex Zhao's proof is incorrect, but I think I've managed to make the idea work.  As in the original, this is just a big angle chase.

Let $K$ be the intersection of $(I)$ and $BC$ closest to $C$.
The triangle $ABK$ is symmetric about $BI$.$\qquad(*)$
Therefore $$
\begin{aligned}
\measuredangle IKB =\measuredangle BAI=\measuredangle IAC&\implies\measuredangle IAC+\measuredangle CKI=\pi\\
&\implies CKIA \text{ is cyclic}\\
&\implies \measuredangle KIA=\pi-\measuredangle ACB.\qquad(**)
\end{aligned}
$$
We claim $\measuredangle KPB+\measuredangle KI'B=\pi$, because
$$\begin{aligned}
\measuredangle KPB 
&\overset{}=\measuredangle APB-\measuredangle APK\\
&= \pi-\measuredangle ACB-\measuredangle KIA/2\\
&=(\pi-\measuredangle ACB)/2 \qquad\text{ (by $(**)$)}\\
&=(\measuredangle CAB+\measuredangle CBA)/2\\
&=\measuredangle IAB+\measuredangle IBA\\
&=\pi-\measuredangle AIB\\
&=\pi-\measuredangle BIK\qquad\text{ (by $(*)$)}\\
&=\pi-\measuredangle KI'B.
\end{aligned}
$$
Thus $PBI'K$ is cyclic.
Therefore
$$
\begin{aligned}
\measuredangle BPI'
&=\measuredangle BKI'\\
&=\measuredangle IKB\\
&=\measuredangle BAI\\
&=\measuredangle BAM\\
&=\measuredangle BPM.
\end{aligned}
$$
Thus $P,I',M$ are collinear and we're done.
A: After a suggestion by @brainjam I found an offical solution:


 This was posted initial and it is not a solution, just partial result if someone find it usefull. 
Let $PM$ meet smaller circle at $D$. Then $AD\bot BC$.


Proof: It is easy to see that angle between altitude from $A$ and angle bisector from is $|\beta-\gamma|$.
 Now we see that $\angle MBC = \angle MAC = \alpha $ so $$\angle DPA = \angle MPA = \angle MBA = \alpha+2\beta$$ From here we get (notice: $\alpha +\beta+\gamma = 90^{\circ}$ and that $ADI$ is isosceles) $$\angle DJA = 360^{\circ}-2(\alpha+2\beta) \implies \angle  DAI = \beta - \gamma$$ and we are done.
Let perpendicular from $I$ to $BC$ meet $PM$ at $J$. If we prove that $BC$ bisect $IJ$ we are done, i.e. $J=I'$. But no idea how to finish.

A: Hint: The line through $P$ and $I$ must intersect the perpendicular bisectors of the triangle; in particular it must intersect the bisector of the side $BC$.
A: Let the circle centered at I intersect BC again at K. I claim that PBI'K is cyclic. Note that
\begin{align*}
\measuredangle KPB 
&=\measuredangle APB-\measuredangle APK\\
& = \measuredangle ACB-\frac{\measuredangle AIK}2\\
&=\measuredangle ACB-\measuredangle AIB\\
&=\measuredangle ACB-(\measuredangle ACB+\measuredangle IAC+\measuredangle CBI)\\
&=\measuredangle CAI+\measuredangle IBC\\
&=\measuredangle IAB+\measuredangle ABI\\
&=\measuredangle AIB\\
&=\measuredangle BIK\\
&=\measuredangle KI'B.
\end{align*}
Therefore
\begin{align*}
\measuredangle BPI'
&=\measuredangle BKI'\\
&=\measuredangle IKB\\
&=\measuredangle BAI\\
&=\measuredangle BAM\\
&=\measuredangle BPM.
\end{align*}
Thus BPI'M are collinear and we're done.
