Proving the parallelity of two lines in a convex Pentagon

Question

Can someone show me how to solve this Geometry Problem? I feel like there is a trick behind it that I don’t see

The convex pentagon $ABCDE$ has a circumference and $\overline{AB} = \overline{BD}$ applies. Let the point $P$ be the intersection of the diagonals $AC$ and $BE$. The straight lines $BC$ and $DE$ intersect at point $Q$. Prove that the line $PQ$ is parallel to the diagonal $AD$.

Answer 1

$AB = BD$ implies that their corresponding circular arcs are equal too, so $EB$ is the angle bisector of $\angle \, AED$ and thus $\angle \, AEB = \angle \, DEB$. Let $R$ be the intersection point of $BE$ and $AD$. Then, after a short angle chasing, triangles $AEP$ and $BEQ$ are similar, so $$\frac{EP}{EQ} = \frac{EA}{EB}$$ and triangles $AER$ and $BED$ are similar, so $$\frac{ER}{ED} = \frac{EA}{EB}$$
Consequently, $$\frac{EP}{EQ} = \frac{EA}{EB} = \frac{ER}{ED}$$ which by the Thales' intercept theorem yields $$RD \, || \, PQ$$ and since $R$ lies on $AD$ you get $$AD \, || \, PQ$$

Answer 2

Observe that, $\triangle AEP\sim \triangle BEQ$ and hence $\angle APE=\angle BQE$. From here, it is seen that quadrilateral $PCQE$ is cyclic.

Thus, $\angle CPQ=\angle CEQ=\angle CED=\angle CAD$ and therefore $PQ\parallel AD$.