Advanced Number Theory · Superchallenge problem

In triangle $ABC$, $AB=13$, $BC=15$, and $CA=14$. The bisector of $\angle A$ meets the circumcircle of the triangle again at $M$. Then $AM=\dfrac{a\sqrt{b}}{c}$, where $a$, $b$, $c$ are positive integers, $\gcd(a,c)=1$, and $b$ is squarefree. Find $a+b+c$.