Show that a triangle is isosceles inside a circle Let $\Gamma$  be a circle. Suppose $\Gamma '$ is another circle whose center lies on $\Gamma $. Let these two circles intersect at $A$ and $B$. Let $P$ be a point on $\Gamma $, and let $PB$ intersect $\Gamma'$ again at $Q$. Show that $\triangle PQA$ is isosceles.
I know my first step is to show that $\triangle ABP$ is isosceles when $P$ would lie on the midpoint between $AB$ , but I am not sure how. 
 A: I am writing about one of the cases here, where $P$ is outside circle $\Gamma'$.
$$\begin{align}\angle APQ =& 180^\circ - \angle AOB\\
\angle AQP =& \frac{1}{2} \angle AOB\\
\angle PAQ =& 180^\circ -(180^\circ - \angle AOB) - \frac{1}{2} \angle AOB\\
=& \angle AQP\\
AP=&QP
\end{align}
$$


By swapping $A$ and $B$, we have another case to prove:
$$\begin{align}
\angle QPA =& 180^\circ - \angle AOB\\
\angle PQA =& 180^\circ - \left(180^\circ-\frac{1}{2}\angle AOB\right)\\
=&\frac{1}{2}\angle AOB\\
\angle PAQ =& \angle PQA\\
AP=QP
\end{align}$$

A: As pointed out, your first step isn't true, so it will be impossible to show.
This question is easily approached by angle chasing, using the properties that were given. Let $O'$ be the center of $\Gamma'$.
Let $ \angle PQA = \alpha $.
What can you say about $\angle AO' B$?
What can you say about $\angle APB$?
Hence show that $\angle PAQ = \alpha$.
Thus the triangle is isosceles.
A: It suffices to show that $\angle QPA=\pi-2\cdot \angle PQA$ which can be accomplished with some angle chasing. In this configuration, $Q$ lies between $P$ and $B$, but angle chasing can be done similarly in other configurations (or with directed angles.)
Let $O$ be the center of $\Gamma'$, so it lies on the circumcircle of $APB$. Then
$$
\angle PQA=\pi-\angle BQA=\pi-(\pi-\frac{1}{2}\angle AOB)=\frac{1}{2}\angle AOB=\frac{1}{2}(\pi-\angle APB)
$$
($\angle BQA=\pi-\frac{1}{2}AOB$ because $O$ is center of $\Gamma'$.) This gives us $\angle APB=\pi-2\cdot \angle APB$ as desired.
If you know about spiral similarity, a more direct method would be to note that $A$ is the spiral similarity center mapping $O_1P$ to $OQ$ (where $O_1$ is the center of $\Gamma$), and thus it also maps $PQ$ to $O_1O$. Thus $\triangle APQ\sim\triangle AO_1O$, which is isosceles because $AO_1=O_1O$ was given.
