Dependency of measure of angles from length of edges in a triangle with excircle

Question

Let $P$ be the point where an excircle is tangent to the edge $AB$ of the triangle $ABC$, where $AC={1\over 3}(AB+BC)$. Prove that $\angle PCA=\angle BCP+\angle ABC$.
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All I could do was play around with the angles, which let to nothing but useless tautologies. Just in case, the problem probably doesn't require mathematical analysis. Thank you

Answer 1

Let excircle touch line $AC$ at $Q$. It is well know (and not difficult to prove-with tangent segments) that $CQ = {a+b+c\over 2}= s$ where $a,b,c$ are the sides $BC,AC,AB$ respectively. Since $4b = a+b+c = 2s$ we have $s=2b$ and so $AQ = b$. Because $AP = AQ =b$ (tangent segments from $A$ on excircle) we conclude that triangle $PAC$ is isosceles and thus the conclusion.