The diagonals of a cyclic quadrilateral $PQRS$ intersect at $U$. The tangent to the circle at $R$ meets $PS$ produced at $T$. If $QR = RS$, prove that

1. Relate the base angles of the isosceles triangle formed by the equal chords.

Since $QR = RS$, $\triangle QRS$ is an isosceles triangle.
Therefore, $\angle RQS = \angle RSQ$ (base angles of isosceles $\triangle$).

2. Apply the alternate segment theorem to relate the tangent angle to the inscribed angle.

Since $RT$ is a tangent to the circle at $R$ and $RS$ is a chord, by the alternate segment theorem,
$\angle SRT = \angle RQS$.

3. Conclude that the lines are parallel using alternate angles.

From the previous steps, we have $\angle SRT = \angle RSQ$.
Since $\angle SRT$ and $\angle RSQ$ are alternate interior angles and they are equal, it follows that $QS \parallel RT$.

1. Show that the first pair of corresponding angles are equal using the alternate segment theorem.

By the alternate segment theorem, the angle between the tangent $RT$ and the chord $RS$ is equal to the angle in the alternate segment subtended by $RS$.
$\angle SRT = \angle SQR$
Since $U$ lies on $QS$, $\angle SQR$ is the same as $\angle UQR$.
Therefore, $\angle SRT = \angle UQR$.

2. Show that the second pair of corresponding angles are equal using parallel lines and angles in the same segment.

From part (a), we know that $QS \parallel RT$. Since $PST$ is a straight line acting as a transversal, the corresponding angles are equal:
$\angle PSQ = \angle STR$
Angles subtended by the same arc $PQ$ at the circumference are equal:
$\angle PRQ = \angle PSQ$
Therefore, $\angle PRQ = \angle STR$.
Since $U$ lies on $PR$, $\angle PRQ$ is the same as $\angle URQ$.
Thus, $\angle URQ = \angle STR$.

3. Conclude that the triangles are similar.

In $\triangle QUR$ and $\triangle RST$:

1. $\angle UQR = \angle SRT$
2. $\angle URQ = \angle STR$

Since two pairs of corresponding angles are equal, $\triangle QUR$ and $\triangle RST$ are similar (AA similarity).