$A$ chord $PQ$ of a circle is parallel to the tangent drawn at a point $R$ of the circle. Prove that $R$ bisects the arc $PRQ$.

(N/A) Given: Chord $PQ$ is parallel to the tangent $MN$ at point $R$.
To prove: $R$ bisects the arc $PRQ$, i.e., arc $PR = \text{arc } RQ$.
Proof:
Let $MN$ be the tangent at $R$. Since $PQ \parallel MN$, the alternate interior angles are equal.
Therefore, $\angle MRP = \angle RPQ$ (Let this be $\angle 1 = \angle 2$).
By the alternate segment theorem, the angle between the tangent and the chord is equal to the angle subtended by the chord in the alternate segment.
Thus, $\angle MRP = \angle RQP$ (Let this be $\angle 1 = \angle 3$).
From the above two equations, we get $\angle 2 = \angle 3$, i.e., $\angle RPQ = \angle RQP$.
In $\triangle PQR$, since $\angle RPQ = \angle RQP$, the sides opposite to these angles are equal.
Therefore, $PR = RQ$.
Since the chords $PR$ and $RQ$ are equal, the arcs subtended by them are also equal.
Hence, arc $PR = \text{arc } RQ$, which means $R$ bisects the arc $PRQ$.