Proving angles in a circle are equal 
$A$, $B$, $R$ and $P$ are four points on a circle with centre $O$.
$A$, $O$, $R$ and $C$ are four points on a different circle.
The two circles intersect at the points $A$ and $R$.
$CPA$, $CRB$ and $AOB$ are straight lines.

Prove that angle $CAB$ = angle $ABC$.

Not really sure how to start here. I am thinking about proving the sides $AC = CB$, but not sure how I can do that. The only striking thing is that $AB$ is a diameter, so angle $APB = 90$. Then letting $PAB = x$, one gets $PBA = 90 - x$ and also $CPB = 90$.
However, can't get much further from here.

Since $AB$ is a diameter, as you've already stated, this means $\measuredangle ARB = 90^{\circ}$ as well. As such, you also have $\measuredangle ARC = 90^{\circ}$. Thus, in the circle on the left side, you have $AC$ is its diameter. This means $\measuredangle COA = 90^{\circ}$ (note you could also get $\measuredangle COA = \measuredangle ARC$ from that both angles subtend the same chord of $AC$) and, thus, $\measuredangle COB = 90^{\circ}$ also.
Since $|OA| = |OB|$, you have with $CO$ being a common side that the Pythagorean theorem in $\triangle COA$ and $\triangle COB$ gives $|CA| = |CB|$. This means $\triangle ABC$ is isosceles so $\measuredangle CAB = \measuredangle ABC$.
Update: Instead of using the Pythagorean theorem, I could've used that side-angle-side (SAS) matches so $\triangle COA \cong \triangle COB$ which directly gives $\measuredangle CAB = \measuredangle ABC$.
