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Let $G=BD\cap AC$.

Claim: $E,G,P$ are collinear.

Proof:

Let $S=EG\cap CD$.

It is well known that $(F,S;D,C)=-1$. ( Read key points please )

Since $\angle DQX=\angle DQY$, hence $FQ$ is the external bisector of $\angle DQC$.

So $\angle FQS=90^\circ$ or $SQ\perp FO$. ( Read key points please )

But by $\textbf{Brocard's theorem}$, we know that $\triangle EFG$ is a self polar triangle or $O$ is the orthocenter of $\triangle EFG$ or $SG\perp FO$.

$\therefore S,G,Q$ are collinear, which implies that $E,G,P$ are collinear .

$\angle EPD=\angle BAD=\angle ECD$. Therefore $E,C,P,D$ are concyclic.

Now, $\angle FPD=\angle EPD$ or $PD$ is the angle bisector of $\angle FPE$ or $\angle FPS$.

Since $(F,S;D,C)=-1$, hence $\angle DPC=90^\circ$.

Therefore, $\angle AEB=\angle DEC=180^\circ - \angle DPC=180^\circ = 90^\circ$