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. Prove that AB || CD.
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. Prove that AB || CD.
ANSWER:
Let us draw BE ⟂ PQ and CF ⟂ RS.
As PQ || RS
So, BE || CF
By laws of reflection we know that,
Angle of incidence = Angle of reflection
Thus, ∠1 = ∠2 and ∠3 = ∠4 --- (i)
also, ∠2 = ∠3 (alternate interior angles because BE || CF and a transversal line BC cuts them at B and C) --- (ii)
From (i) and (ii),
∠1 + ∠2 = ∠3 + ∠4
⇒ ∠ABC = ∠DCB
⇒ AB || CD (alternate interior angles are equal)