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$is parallel to $ CD.$

Answer 1

Step 1: Construction.

Draw two lines $BE$and $CF$such that $BE\perp PQ$ and $CF\perp RS.$

Now, $BE\parallel CF$ and $BC$ is a transversal.

$\angle EBC=\angle FCB$ [Alternate interior angles] ……$\left[1\right]$

Step 2: Prove that $AB$ is parallel to $CD.$

$\angle ABE=\angle EBC$and $\angle FCB=\angle FCD$ (Since, angle of incidence $=$angle of reflection)

$\Rightarrow \angle ABE=\angle FCD$ ( from equation $\left[1\right]$) ……$\left[2\right]$

On adding equations $\left[1\right]$ and $\left[2\right]$ we get,

$$\begin{gathered} \angle EBC+\angle ABE=\angle FCB+\angle FCD \\ \Rightarrow \angle ABC=\angle BCD \end{gathered}$$

i.e. a pair of alternate angles are equal.

Hence proved that $AB\parallel CD.$