2019 AIME #13 (problem request)

This problem was requested to be solved by a user. If you would like to request a problem too, please use the form on the left menu of the blog. 

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 $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, 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 $a+b+c+d$. 

Let $P=AE\cap CF$, $CP=5x$, and $BP=5y$. By $\triangle{CBP}\sim\triangle{EFP}$, we know that $EP=7x$ and $FP=7y$. By $\triangle{CAP}\sim\triangle{DFP}$ we know that $$\frac{6}{4+5y}=\frac{2}{7y} \implies y=\frac{1}{4}.$$ Hence $BP=\frac{5}{4}$ and $FP=\frac{7}{4}$. Note that the similar triangles also give $DP=\frac{5}{3}x$ so $DE=\frac{16}{3}x$. Then, by Stewart's Theorem on $\triangle{FEP}$ and cevian $FD$, we obtain \[\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x \implies x=\frac{3\sqrt{2}}{4}.\] Finally, we get $BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}$ so the answer is $\boxed{032}$ as desired.