Saudi Arabian Mathematical Competitions

Because triangles $PBC$ and $PAE$ are similar, and triangles $QED$ and $QAB$ are similar, we have
$$\frac{BC}{AE} = \frac{PB}{PA}, \quad \text{and} \quad \frac{DE}{BA} = \frac{EQ}{AQ}.$$

Therefore
$$\begin{equation*} \frac{BC \cdot DE}{AE \cdot BA} = \frac{PB \cdot EQ}{PA \cdot AQ} = \frac{PQ^{2}}{PA \cdot AQ}. \tag{*} \end{equation*}$$
On the other hand, applying sine law on triangles $ABQ$ and $APQ$, we have
$$\frac{AB}{AQ} = \frac{\sin \angle AQB}{\sin \angle EBA}, \quad \text{and} \quad \frac{PQ}{AP} = \frac{\sin \angle PAQ}{\sin \angle AQP}.$$
Finally, by applying sine law on the circumcircle of the pentagon $ABCDE$, we get
$$\frac{AB}{AQ} \cdot \frac{PQ}{AP} = \frac{\sin \angle AQB}{\sin \angle EBA} \cdot \frac{\sin \angle PAQ}{\sin \angle AQP} = \frac{\sin \angle CAD}{\sin \angle EBA} = \frac{CD}{EA}. \quad \text{(**)}$$
Combining the two relations ($*$) and ($**$), we obtain after cancellation
$$BC \cdot DE = CD \cdot PQ.$$