Question

In the 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 $AB\parallel CD$

Answer 1

From laws of refection $\angle 1=\angle 2and\angle 3=\angle 4$ (IMAGE)

So, $\angle 1=\angle 2=\frac{1}{2}\angle ABCand\angle 3=\angle 4=\frac{1}{2}\angle BCD$

Here BM and CN are Normal

So $BM\perp PQandCN\perp RS$

put,$PQ\parallel RS$

So, $BM\parallel CN$

Now $BM\parallel CN$, and $BC$ is transversal

$\Rightarrow \angle 2=\angle 3$

$\Rightarrow \frac{1}{2}\angle ABC=\frac{1}{2}\angle DCB$

$\Rightarrow \angle ABC=\angle BCD$

$AB\parallel CD$ as it transversal $BC$

Intersect two lines such that pair of interior angle is equal then lines are parallel.