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 that AB || CD.

Answer 1

Given: PQ || RS and AB is the angle of incident, CD is the reflected ray.

To prove: AB || CD

Construction: Draw BM ⊥ PQ and CN ⊥ RS

Proof: BM ⊥ PQ and CN ⊥ RS

∴ BM || CM and BC is a transversal.

∴ ∠2 = ∠3            ...(1)        ...(alternate interior angles)

While we know that

The angle of incident = Angle of reflection, where BM and CN are perpendicular.

∴ ​∠1 = ∠2           ...(2)

Similarly,

∴ ​∠3 = ∠4          ...(3)

From Equations (1), (2) and (3) we get

∠1 = ∠4              ...(4)

Addition of Equations (1) and (4)

∠1 + ∠2 = ∠3 + ∠4

∠ABC = ∠BCD         ...(alternate interior angles)

Therefore, AB || CD