In Fig. 5.63, exa

i) ∠ CQF≠∠ SRD (Alternate Exterior angle)

Hence EF is not || GH

ii) ∠ CQF = ∠ RQF = 65⁰ (Vertically Opposite)

∠ RQF + ∠ SPQ = 115⁰ + 65⁰ = 180⁰

Hence they form a pair of corresponding interior angle. So AB || CD