Japan Junior Mathematical Olympiad

$$\boxed{\frac{24}{5}}$$ Let $F$ be the point of intersection of $BD$ and $EC$. From $AB//EF$, we get $AB:EF=DA:DE$. Therefore, we have $EF=\frac{AB\cdot DE}{DA}=\frac{9}{5}$.

We also have $AB=CB=3$, $AD=CD=5$, which imply that the triangles $ABD$ and $CBD$ are congruent as the side $BD$ is common to both. Consequently, we have $\angle ABD=\angle CBD$, and this, together with the fact $AB//EC$ yields $\angle CFB=\angle ABF=\angle CBF$. Therefore, we have $FC=BC=3$. We finally obtain $EC=EF+FC=\frac{24}{5}$ for the desired answer.