Question

In 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.​

Answer 1

First, draw two lines $BE$ and $CF$ such that $BE\perp PQ$ and $CF\perp RS.$


Now, since $PQ||RS$,
So, $BE||CF$
We know that,
Angle of incident = Angle of reflection
$\therefore \angle 1=\angle 2$ and $\angle 3=\angle 4$

$BE||CF$ and $BC$ is the transversal.
So, $\angle 2=\angle 3$ (Alternate interior angles)
$\therefore \angle 1=\angle 2=\angle 3=\angle 4$

Now, $\angle 1+\angle 2=\angle 3+\angle 4$
or, $\angle ABC=\angle DCB$
So, $AB||CD$ (Alternate interior angles are equal)