Mediterranean Mathematical Competition

From the similarity of triangles $BEC$ and $AED$ we $BE=42$.

We put $DE=x$ and $CE=y$. From the similarity of triangles $ABE$ and $DEC$ we have $$\frac{x}{AE}=\frac{y}{BE}=\frac{CD}{AB}\iff \frac{x}{15}=\frac{y}{14}=\frac{CD}{13}=t.(1)$$ Moreover, from theorem of Ptolemy we find $$\begin{array}{rl}BD\cdot AC& =AD\cdot BC+AB\cdot CD\\ \iff (45+14t)(42+15t)=60\cdot 56+39\cdot 13t& \\ \iff 35t^2+126t-245=0\iff t=\frac{7}{5}\text{ or }t=-5.& \end{array}$$ Since $t$ must be positive, from (1) we have $t=\frac{7}{5}$ and $CD=\frac{91}{5}$.