smc

Let $X$ be a point on $BC$ so $AX\perp BC$. Let $AX=h$, $EX=\sqrt{64-h^2}$ and $BX=\sqrt{144-h^2}$. $CE=CX-EX=\sqrt{144-h^2}-\sqrt{64-h^2}$. Using Power of a Point on $E$, $(BE)(EC)=(AE)(ED)$ (there isn't much information about the circle so I wanted to use PoP). $$(\sqrt{144-h^2}-\sqrt{64-h^2})(\sqrt{144-h^2}+\sqrt{64-h^2})=8(ED)$$ $$(144-h^2)-(64-h^2)=8(ED)$$ $$80=8(ED)$$ $$ED=10$$ Adding up $AD$ and $ED$ we get $\boxed{18}$.