Answer: One of seg TR, seg TS cuts the circle at a point Q; \(\angle RQS = 90^\circ\) is an exterior angle of a triangle with \(\angle RTS\) as a remote interior angle, so \(\angle RTS < 90^\circ\).
T is outside the circle, so at least one of the segments TS, TR crosses the circle at a second point. Suppose seg TS meets the circle at Q (T–Q–S); the other case is the same with R and S interchanged.
Draw seg RQ.
- \(\angle RQS = 90^\circ\) (angle inscribed in a semicircle, RS is a diameter).
- \(\angle RQS\) is an exterior angle of \(\triangle RQT\), so \(\angle RQS = \angle QTR + \angle QRT\).
Hence \(\angle QTR < \angle RQS = 90^\circ\).
But \(\angle QTR\) is the same as \(\angle RTS\) (Q lies on seg TS). \(\therefore \angle RTS < 90^\circ\), i.e. \(\angle RTS\) is an acute angle.