In the above figure, chord PQ and chord RS intersect at point T. Then complete the following activity to prove, \(\angle STQ = \dfrac{1}{2}\left[m(\text{arc } SQ) + m(\text{arc } PR)\right]\)[Figure: a circle with points P, S above and R, Q below on…

Answer: the blanks are \(\angle PQR\), \(\frac{1}{2}m(\text{arc } PR)\), \(m(\text{arc } SQ)\) and \(m(\text{arc } PR)\).

\(\angle STQ\) is an exterior angle of \(\Delta QTR\) (ST is the extension of RT), so \(\angle STQ = \angle TQR + \angle TRQ = \angle PQR + \angle SRQ\).

Now \(\angle SRQ\) and \(\angle SPQ\) are inscribed angles in the same segment, so \(\angle SRQ = \angle SPQ\). Hence \(\angle STQ = \angle SPQ + \angle PQR\).

By the inscribed angle theorem, \(\angle SPQ = \frac{1}{2}m(\text{arc } SQ)\) and \(\angle PQR = \frac{1}{2}m(\text{arc } PR)\).

\(\therefore \angle STQ = \frac{1}{2}m(\text{arc } SQ) + \frac{1}{2}m(\text{arc } PR) = \frac{1}{2}\left[m(\text{arc } SQ) + m(\text{arc } PR)\right]\), as required.