A chord \(PQ\) of a circle is parallel to the tangent at a point \(R\) of the circle. Prove that \(R\) bisects the arc \(PRQ\).
\(\text{Arc }PR=RQ\) (so \(R\) bisects arc \(PRQ\)).
Step 1: Draw a circle with center \(O\). Mark chord \(PQ\) and a tangent at point \(R\). Given: \(PQ \parallel \text{tangent at } R\).
Step 2: Since the chord \(PQ\) is parallel to the tangent at \(R\), the angle made at \(R\) by line \(RP\) (say \(\angle QRP\)) will be equal to the angle at \(P\) inside the triangle (\(\angle QPR\)). These are alternate interior angles from parallel lines.
Step 3: Now apply the Tangent–Chord Theorem: The angle between a tangent and a chord through the point of contact is equal to the angle in the opposite arc. So, \(\angle QRP = \angle PQR\).
Step 4: From Step 2, we already have \(\angle QRP = \angle QPR\). From Step 3, we have \(\angle QRP = \angle PQR\). Therefore, \(\angle QPR = \angle PQR\).
Step 5: This means triangle \(PQR\) is isosceles (two angles equal). So, arc \(PR\) = arc \(RQ\).
Final Step: Since both arcs are equal, point \(R\) divides the bigger arc \(PRQ\) into two equal parts. Hence, \(R\) bisects arc \(PRQ\).