First, draw two lines BE and CF such that BE⊥PQ and CF⊥RS .Now, since PQ II RS,So, BE II CFWe 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 B C cuts them at B and C .So, ∠2=∠3 (As they are alternate interior angles)Now, ∠1+∠2=∠3+∠4 Or, ∠ABC=∠DCB So, A B II C D alternate interior angles are equal)