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There is given a trapezoid $ABCD$ in the plane with $BC \parallel AD$. We know that the angle bisectors of the angles of the trapezoid are concurrent at $O$. Let $T$ be the intersection of the diagonals $AC$ and $BD$. Let $Q$ be the foot of the altitude from $O$ to $CD$. Prove that if the circumcircle of the triangle $OTQ$ intersects $CD$ again at $P$, then $TP \parallel AD$.

The angle bisectors of the angles of the trapezoid $ABCD$ are concurrent at $O$, which implies that there exists a circle $\Gamma$ centered at $O$ inscribed in trapezoid $ABCD$. Let $Q_1$ and $Q_2$ be the touchpoints of $\Gamma$ with $BC$ and $AD$, respectively. We prove the following two claims.

$\textbf{Claim 1:}$ $T$ lies on $Q_1Q_2$.
$\textbf{Proof:}$ Apply Brianchon's Theorem to the degenerate hexagon $AQ_2DCQ_1B$.

$\textbf{Claim 2:}$ $O$ lies on $Q_1Q_2$.
$\textbf{Proof:}$ Let $OQ_1$ intersect $AD$ at $Q'_2$. As $OQ_1 \perp BC$, $BC \parallel AD$, we have $OQ'_2 \perp AD$. This implies $Q_2 = Q'_2$, as desired.

Let $P'$ be a point on $CD$ such that $TP' \parallel AD \parallel BC$. Now, $\measuredangle OQP' = \measuredangle OQC = \measuredangle OQ_1C = \measuredangle OTP' = 90^o$. Therefore, $O, Q, P'$ and $T$ are concyclic. Thus, $P = P'$, as desired.