# Collinear Proof in Circle Geometry

(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 on 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. 