At point B, draw BM ⊥ RS and at point C, draw CN ⊥ PQ
∠1 = ∠2 …(i)
and ∠3 = ∠4 ,..(ii)
(angle of incidence is equal to angle of reflection)
Also ∠2 = ∠3 …(iii)
(Alternate angles)
From (i), (ii) and (iii), we get
⇒ ∠1 = ∠4
⇒ 2∠1 = 2∠4
⇒ ∠1 + ∠1 = ∠4 + ∠4
⇒ ∠1 + ∠2 = ∠3 + ∠4
[using (i) and (ii)]
⇒ ∠ABC = ∠BCD
(Alternate angles)
⇒ AB || CD