Question

In fig.6.33, 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.

$\bf{Prove \: that\: AB \:is \:parallel\: to \:CD.}$

class - IX

Need well explained answer.

Answer 1

here is your answer OK Di.......


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 cuts 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

Answer 2

this is your Answer thanks