Problem 1483

1. Identify the key points and properties:
- Given trapezoid $ABCD$ with $AD \parallel BC$ and $AD \perp AB$.
- $T$ is the intersection of diagonals $AC$ and $BD$.
- A circle centered at $O$ is inscribed in the trapezoid and touches $CD$ at point $Q$.
- $P$ is the intersection point (different from $Q$) of the side $CD$ and the circle passing through $T, Q$, and $O$.

2. Define the contact points:
- Let $E$ and $F$ be the contact points of the circle with lines $AD$ and $BC$ respectively.

3. **Collinearity of $E, O, F$:**
- Since $\angle DAB = \angle ABC = 90^\circ$, it is easy to see that $E, O, F$ are collinear. This is because the tangents from a point to a circle are equal, and the perpendiculars from the points of tangency to the center of the circle are radii of the circle.

4. Application of Brianchon's Theorem:
- By Brianchon's theorem in the hexagon $AEDCFB$, the diagonals $AE, CF, BD$ must be concurrent. Since $T$ is the intersection of $AC$ and $BD$, and $E, O, F$ are collinear, it follows that $E, T, O, F$ are all on the same line.

5. Perpendicularity and angles:
- Since $OQ$ is the radius of the circle and touches $CD$ at $Q$, $OQ \perp CD$.
- Therefore, $\angle OQP = 90^\circ$.

6. Angle relationships:
- Since $P$ lies on the circle passing through $T, Q$, and $O$, $\angle OTP = 90^\circ$.
- Also, $\angle OED = 90^\circ$ because $OQ$ is perpendicular to $CD$.

7. Conclusion:
- Since $\angle OTP = \angle OED = 90^\circ$, it follows that $TP \parallel AD$.

$\blacksquare$