• $FQ$ is the external bisector of $\angle DQC$.
• For some collinear points $A,B,C,D$, The expression $(A,B;C,D)$ represents the value of $\frac{AC}{AD} \times \frac{BD}{BC}$, calculated using directed lengths. To understand this concept better, consider $(F,S;D,C)$. Here $SD/SC$ is negative because $SD$ and $SC$ point in opposite directions. However, $SD/CS$ is positive, as swapping the labels changes the direction of comparison. For more details, refer to Chapter $9$ of $\textit{EGMO}$.
• To know why $(F,S;D,C)=-1$, read chapter $9$ of $\textit{EGMO}$.
• $E,S,G,P,Q$ are collinear.
• By \textbf{Brocard's theorem}, $O$ is the orthocenter of $\triangle EFG$
• Using Lemma $9.18$ from EGMO, we deduced that $FQS=90^\circ$ or $SQ\perp FO$. Additionally, the same lemma is used to deduce that $\angle DPC=90^\circ$.
• $D,P,C,E$ are concyclic.

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