6. 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.
Solution: First, draw two lines BE and CF such that BE ⊥ PQ and CF ⊥ RS. Now, since PQ || RS, So, BE || CF We know that, Angle of incidence = Angle of reflection (By the law of reflection) So, ∠1 = ∠2 and ∠3 = ∠4 We also know that alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at B and C So, ∠2 = ∠3 (As they are alternate interior angles) Now, ∠1 +∠2 = ∠3 +∠4 Or, ∠ABC = ∠DCB So, AB || CD (alternate interior angles are equal)
