Question 58
Let $AEF$ be a triangle with $EF=20$ and $AE=AF=21$. Let $B$ and $D$ be points chosen on segments $AE$ and $AF$, respectively, such that $BD$ is parallel to $EF$. Point $C$ is chosen in the interior of triangle $AEF$ such that $ABCD$ is cyclic. If $BC=3$ and $CD=4$, then the ratio of areas $\frac{[ABCD]}{[AEF]}$ can be written as $\frac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100a+b$.