Circle $\Gamma$ with center $O$ is tangential to $PA$ at $A$. Line $PO$ extended intersects $\Gamma$ at $B$ (i.e. $PO < PB$). $C$ is a point on $AB$ such that $PC$ bisects $ \angle APB$. What is the measure (in degrees) of $ \angle PCA$?
+128460 
Let AP= AO
Angle PAO = 90
So angle APO =Angle AOP = 45
Then since PC is an angle bisector, then Angle APC = 22.5
And Angle AOP = Angle AOD = 45
The angle ABD is an inscribed angle intercepting an arc of 45 ....so its measure = 22.5 = angle OBA
And in triangle AOB, OA = OB....so Angle OBA = Angle OAB = 22.5
And angle CAP = Angle PAO + Angle OAB = 90 + 22.5 = 112.5
So in triangle APC, angle PCA = 180 - 112.5 - 22.5 = 180 - 135 = 45 (degrees)
