$\triangle BEC\sim\triangle AED$ so $BE=42$. Using Stewarts theorem on $\triangle ABD$, solving for $DE=21$. From $\triangle AEB\sim\triangle DEC$, we get $CD=\frac{91}{5}$.

It would actually be easier if you just found ED,EC in terms of CD and then used Ptolemy's Theorem on ABCD.

Triangles AED and BEC are similar, then easily BE=42. Triangles AEB and DEC are similar, then for some real k, DE=45k, EC=42k and CD=39k. Now apply Ptolemy's theorem and solve for k (you obtain a quadratic equation).