Question

In Fig. 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.

Answer 1

Draw $BL\perp PQ$ and $CM\perp RS$. As $PQ\parallel RS$, so, $BL\parallel CM$.

The alternate interior angles are equal, so,

$\angle LBC=\angle MCB$ (1)

It is known that the angle of reflection is equal to the angle of incidence, therefore,

$\angle ABL=\angle LBC$ (2)

And,

$\angle MCB=\angle MCD$ (3)

From equation (1), (2) and (3),

$\angle ABL=\angle MCD$ (4)

Add equation (1) and (4),

$\angle LBC+\angle ABL=\angle MCB+\angle MCD$

$\angle ABC=\angle BCD$

Since, these are the interior angles and are equal, hence, $AB\parallel CD$.