The diagram shows a circle passing through the points P, Q, R and S. Chord SQ and RP intersect at T. The straight line URV is a tangent to the circle at R. The tangent meets the line SQ extended at U such that RU = RS.

1. Use the alternate segment theorem to relate angle URQ to angle QSR.

By the alternate segment theorem, $\angle URQ = \angle QSR$.

2. Use the result from part (a) to relate angle URQ to angle RQU, and then to angle QSR.

From part (a), $\triangle RQU$ is isosceles with $RU = RQ$, so $\angle URQ = \angle RQU$. Therefore, $\angle RQU = \angle QSR$.

3. Use angles in the same segment to relate angle QSR to angle QPR, and angle RQS to angle RPS.

Angles in the same segment are equal, so $\angle QPR = \angle QSR$ and $\angle RPS = \angle RQS$.

4. Combine the equalities to show the required relationship.

Since $\angle RQU = \angle RQS$ (same angle) and $\angle RQU = \angle QSR$, we have $\angle RQS = \angle QSR$.
Thus, $\angle RPS = \angle QPR$.
Since $T$ lies on $RP$, $\angle QPT = \angle QPR$ and $\angle SPT = \angle RPS + \angle QPR = 2 \angle QPR = 2 \angle QPT$.