smc

Let $BE=a$, $AD=b$, and $AC=CE=BD=c$. Let $F$ be on $AE$ such that $CF\perp AE$. In $△CFE$ we have $\cos \theta =-\cos (\pi -\theta )=-7/c$. We use the Law of Cosines on $△ABC$ to get $60\cos \theta =109-c^2$. Eliminating $\cos \theta$ we get $c^3-109c-420=0$ which factorizes as $$(c+7)(c+5)(c-12)=0.$$Discarding the negative roots we have $c=12$. Thus $BD=AC=CE=12$. For $BE=a$, we use Ptolemy's theorem on cyclic quadrilateral $ABCE$ to get $a=44/3$. For $AD=b$, we use Ptolemy's theorem on cyclic quadrilateral $ACDE$ to get $b=27/2$. The sum of the lengths of the diagonals is $12+12+12+\frac{44}{3}+\frac{27}{2}=\frac{385}{6}$ so the answer is $385+6=\boxed{391}$