Question

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 $AB\parallel CD$.

Answer 1

At surface $B,$ $\angle 1=\angle 2$ (property of reflection)

At surface $C,$ $\angle 3=\angle 4$ (property of reflection))

As the sum of the angles are equal $AB$ is parallel to $CD,$

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

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

$\because$Alternate interior angles are equal.

$\therefore AB\parallel CD$.