jmc

Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F\ne C$ is a point of intersection of the circumcircles of $△ACD$ and $△EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\frac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.