In the diagram, PQT is a straight line and SQ // RT. (a) Join QR and show that : (i) < RPS = < QRT ; (ii) < PRS = < QTR. (b) ABC is a triangle. The sides AB...

(a) P, Q, R, S lie on the circle, PQT is a straight line (T on PQ produced beyond Q), and \(SQ\parallel RT\). Join QR.

(i) Show that \(\angle RPS=\angle QRT\).

• \(\angle RPS\) and \(\angle RQS\) both stand on the same chord \(RS\) and P, Q are in the same segment, so \(\angle RPS=\angle RQS\) (angles in the same segment).
• \(SQ\parallel RT\) with QR as transversal gives \(\angle RQS=\angle QRT\) (alternate angles).

Therefore \(\angle RPS=\angle QRT\), as required.

(ii) Show that \(\angle PRS=\angle QTR\).

• \(\angle PRS\) and \(\angle PQS\) both stand on the same chord \(PS\) with R, Q in the same segment, so \(\angle PRS=\angle PQS\) (angles in the same segment).
• The straight line PQT is a transversal cutting the parallel lines \(SQ\) and \(RT\) at Q and T, so \(\angle PQS=\angle QTR\) (corresponding angles).

Therefore \(\angle PRS=\angle QTR\), as required.

(b) In triangle ABC, AB is produced to D and AC is produced to E, with \(\angle DBC=132^\circ\) and the exterior angle at C equal to \(96^\circ\).

\(\angle DBC\) is the exterior angle at B, and it is the supplement of the interior angle \(\angle ABC\) (angles on the straight line ABD):

The exterior angle at C is the supplement of the interior angle \(\angle ACB\) (angles on the straight line ACE):

The three interior angles of triangle ABC sum to \(180^\circ\):

Hence \(\angle BAC=\angle ABC=48^\circ\). Since two angles are equal, the sides opposite them are equal, i.e. \(BC=AC\). Therefore triangle ABC is isosceles.