This picture is the second question, right? I can only answer so much. Because AF=CD, AC = DF is deduced; because AB//DE, ∠ A = ∠ D is deduced; Because AB=DE, AC=DF, ∠A=∠D, it is concluded that triangle ABC is congruent with triangle DCE. So we can deduce that ∠ACB=∠EFD, so BC//EF.