Supose ABC is that triangle. Let $\Sigma$ the elipse of the points $Q$ such that $AQ+QB=AC+BC$, and $\Gamma$ the circle of centre $P$ and radius $7$. $C$ is the unique intersection point of $\Sigma$ and $\Gamma$ ( because the choice of $ABC$). In particular , there is a comun tangent $\iota$ to $\Sigma$ and $\Gamma$ passing by $C$. Let $X$ and $Y$ points in $\iota$ such that $C$ is betwen them. For property of the elipse we have $<XCA=<YCB$. Since $\iota$ is tangent to $\Gamma$ then $<ACP=<BCP$.

And what about this similar problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=116987 :

here is a very similar problem. http://www.mathlinks.ro/Forum/viewtopic.php?t=114926

Yes.. This makes you wonder -- maybe there is a generalization for this...

Yes, it can be generalized. It's not hard to see that the same procedure works for any fixed distances $ PA, PB, PC $.