(a) \( \ \ \) Show that \( \angle DSR = \angle DAR \).
The quadrilateral \( DRAS \) is cyclic, since \( \angle DRA + \angle DSA = \pi (180^{\circ}) \).
Therefore \( \angle DSR = \angle DAR \) (angles in the same segment of circle \( DRAS \), both standing on same chord \( DR \).
(b) \( \ \ \) Show that \( \angle DST = \pi – \angle DCT \).
Since \( \angle DSC = \angle DTC \) both \( \displaystyle \frac{\pi}{2} \) and these stand on the same interval \( DC \) and the same side of it, then \( DSTC \) is a cyclic quadrilateral.
Thus \( \angle DST + \angle DCT = \pi \) (opposite angles of cyclic quadrilateral \( DSTC \))
\( \therefore \angle DST = \pi-\angle DCT \)
(c) \( \ \ \) Deduce that the points \( R, S \) and \( T \) are collinear.
\( \angle DSR + \angle DST = \angle DAR + \pi – \angle DCT \cdots (1) \)
\( \angle DAR = \angle DCB \) (exterior angle of cyclic quadrilateral \( DABC \) equals the interior opposite angle).
\( \angle DCB = \angle DCT \) (same angle)
\( \angle DAR = \angle DCT \)
\( \angle DSR + \angle DST = \angle DCT + \pi-\angle DCT \cdots (1) \)
\( \angle DSR + \angle DST = \pi \)
Thus \( \angle RST \) is a straight angle, and therefore points \( R,S \), and \( T \) are collinear.
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