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 || CD.

Draw perpendiculars BE and CF on PQ and RS, respectively.

Therefore, we can say that BE || CF
As angle of incidence = angle of reflection,
we get, \(\angle{a}\) = \(\angle{b}\) ....(i)
Similarly, we get, \(\angle{x}\) = \(\angle{y}\) ....(ii)

By Alternate angles theorem, \(\angle{b}\) = \(\angle{y}\)
Now, doubling the angles, we get,
2\(\angle{b}\) = 2\(\angle{y}\)
\(\Rightarrow \) \(\angle{b}\) + \(\angle{b}\) = \(\angle{y}\) + \(\angle{y}\)
from (i) and (ii),
\(\angle{a}\) + \(\angle{b}\) = \(\angle{x}\) + \(\angle{y}\)

Hence, \(\angle{ABC}\) = \(\angle{DCB}\)

Thus by converse of alternate angles theorem,

we get, AB || CD
Hence, it is proved.