In the given 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 || CD.

Solution:

Let us draw $B M \perp P Q$ and $C N \perp R S$.

As PQ || RS,

Therefore, BM || CN

Thus, $B M$ and $C N$ are two parallel lines and a transversal line $B C$ cuts them at $B$ and $C$ respectively.

$\therefore \angle 2=\angle 3$ (Alternate interior angles)

Also, $\angle 1+\angle 2=\angle 3+\angle 4$

$\angle \mathrm{ABC}=\angle \mathrm{DCB}$

However, these are alternate interior angles.

$\therefore \mathrm{AB} \| \mathrm{CD}$
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