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 $||$ CD.

Answer 1

Let us draw BM $⊥$ PQ and CN $⊥$ RS.
As PQ $||$ RS,
Therefore, BM $||$ CN
Thus, BM and CN are two parallel lines and a transversal line BC cut them at B and C respectively.
$∴$ $∠$2 = $∠$3 (Alternate interior angles)
However, $∠$1 = $∠$2 and $∠$3 = $∠$4 (By laws of reflection)
∴ $∠$1 = $∠$2 = $∠$3 = $∠$4
Also, $∠$1 + $∠$2 = $∠$3 + $∠$4
$∠$ABC = $∠$DCB
However, these are alternate interior angles.
∴ AB $||$CD