TITLE: A geometry problem with the reflection of the incenter
QUESTION [5 upvotes]: 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:

REPLY [3 votes]: @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.