In the figure,$PQ$ and $RS$ are two mirrors placed parallel to each other. An incident ray $AB$ strikes the mirror $PQ$ at $B$,the reflected ray moves along the path $BC$ and strikes the mirror $RS$ at $C$ and again reflects back along $CD$. Prove that $AB \parallel CD$.

(N/A) $(i)$ Perpendiculars to parallel lines are parallel.
$(ii)$ According to the laws of reflection,the angle of incidence $=$ the angle of reflection.
Draw ray $BL \perp PQ$ and $CM \perp RS$.
$\because PQ \parallel RS$ [Given]
$\therefore BL \parallel CM$ [$\because BL \perp PQ$ and $CM \perp RS$]
And $BC$ is a transversal.
$\therefore \angle LBC = \angle MCB$ [Interior alternate angles] ......... $(1)$
Since,(Angle of incidence) $=$ (Angle of reflection)
$\therefore \angle ABL = \angle LBC$ and $\angle MCB = \angle MCD$
$\Rightarrow \angle ABL = \angle MCD$
$\therefore$ From $(1)$,we have
$\angle LBC + \angle ABL = \angle MCB + \angle MCD$ [Equals are added to equals]
$\Rightarrow \angle ABC = \angle BCD$
i.e.,the angles of a pair of interior alternate angles are equal.
$\therefore AB \parallel CD$.