Prove that the line segment joining the points of contact of two parallel tangents to a circle passes through its…

Sol.
QQ' $||$ PP'
Let P and Q be the points of contact and O is the centre of the circle.
Join OP and OQ. Draw OA $||$ QQ'.
$\therefore$ QQ' $\perp$ OQ $\Rightarrow \angle 1 = 90^{\circ} \Rightarrow \angle 2 = 90^{\circ}$ ----- (i)
Since OQ' $||$ PP'
$\therefore$ OA $||$ PP' and hence $\angle 4 = 90^{\circ}$ ----- (ii)
Adding (i) and (ii),
$\angle 2 + \angle 4 = 180^{\circ}$ or $\angle POQ = 180^{\circ}$
$\therefore$ POQ is a straight line.