Construction: Draw perpendicular line from D and C to AB such that it cuts AB at F and E, respectively.

Now, In ∆​ADF and ∆BCE,

AD = BC (Given)

∠AFD = ∠BEC (90^o each)

DF = CE (Perpendicular distance between the same parallels)

∴ By SSA axiom

∆​ADF ≅ ∆BCE​

∠A​ = ∠B (by c.p.c.t.)

Therefore Option A is correct.