Prove that triangle RVU is similar to triangle RPQ.

QR is the diameter of the circle. The angle subtended by a diameter at any point on the circumference is $90^\circ$.
Therefore, $\angle RPQ = 90^\circ$.
It is given that UV $\perp$ QR.
Therefore, $\angle RVU = 90^\circ$.

Both triangles, $\triangle RVU$ and $\triangle RPQ$, share the angle at vertex R.
Thus, $\angle VRU = \angle PRQ$ (common angle).

Since two angles of $\triangle RVU$ are equal to two corresponding angles of $\triangle RPQ$ ($\angle RVU = \angle RPQ = 90^\circ$ and $\angle VRU = \angle PRQ$), the triangles are similar by the Angle-Angle (AA) similarity criterion.
Therefore, $\triangle RVU ||| \triangle RPQ$.