Question

In 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

We draw $BE\perp RS$, then BE is also $\perp PQ(\because PQ\parallel RS)$
We draw $CF\perp PQ$. Here also $CF\perp RS$
Here, if we consider PQ as transversal intersecting lines BE and CF, then each pair of corresponding angles is equal. (each equal to ${90}^o$)
Thus we have $BE\parallel CF$.
Now, $\angle ABE=\angle CBE$
(Angle of incidence $=$ Angle of reflection)
$\Rightarrow \angle ABE=\angle CBE=\frac{1}{2}\times \angle ABC$ .....(1)
Similarly, $\angle BCF=\angle FCD=\frac{1}{2}\times \angle DCB$ ....(2)
Now, $BE\parallel CF$
$\Rightarrow \angle CBE=\angle BCF$ (alternate angles)
$\Rightarrow \frac{1}{2}\times \angle ABC=\frac{1}{2}\times \angle DCB$ [by (1) and (2)]
$\Rightarrow \angle ABC=\angle DCB$
$\Rightarrow AB\parallel CD$