Excircle and circumcircle $$ \triangle ABC$$
$k(O; R)$ circumscribed, $P$ is the midpoint of arc $AB$
$O_1$ is the centre of the excircle of $\triangle ABC$, touching the side AB.
Show that $PA = PB = PO_1$.

I tried to prove that $\triangle APO_1$ (or $\triangle BPO_1$) is isosceles, but I can't figure it out. I will be very grateful if you help me.

 A: Let I, J be the in-center and ex-center respectively.

Note that C, I, P are collinear and C, I, J are also collinear. Then, CIPJ are on the same straight line.
"$\angle AIP = ... = \angle ABJ$" implies AIBJ is cyclic.
"$IB \bot JB$" implies IPJ is the diameter of the circle AIBJ.
The perpendicular bisector of chord AB cuts CPJ at P. This means P is the center of AIBJ.
A: Solving this can be divided into 2 stages:


*

*Show that $P$ lies on $CO_1$. 

*Angle chasing
For the first stage, connect $CO_1$, name the intersection of $CO_1$ and arc $AB$ be $P'$. Apparently, $CO_1$ is the bisector of $∠MCN$, so $∠ACP'=∠P'CB$. Also, since $∠ACP=∠PCB$, we can deduce that $P'$ overlaps with $P$. So $P$ is on $CO_1$, and $PA=PB$.
For the second stage, start from $∠MAP=∠CBP$. With $∠MAO_1=∠QAO_1$, $∠PAB=∠PBA$, we can rewrite the equality to $∠QAO_1+∠PAO_1=∠CBA+∠PAB$, which can be reduced to $2∠PAO_1=∠CBA$. And since $∠CBA=∠CPA$, we can then deduce $∠CPA=2∠PAO_1$, which is the same as saying $∠PAO_1=∠PO_1A$. Therefore $PA=PO_1$.
Hence $PA=PB=PO_1$ is shown.
