help geometry

Since $\overline{AB}\parallel \overline{DE}$, we have $\triangle BCD \sim \triangle ECD$ by AA Similarity. Since $BD=24$, $BC:CD=24:CD$ or equivalently $BC:CD=CD:24$. Thus $\triangle BCD \sim \triangle ECD \sim \triangle BDC$. In particular, $\angle CDB \cong \angle CBD$, so $\triangle BCD$ is isosceles with $CD=BD=\boxed{24}$.