As shown in the figure, points B, D, C, f are on a straight line, ab = De, angle a = angle D, AC is parallel to DF, prove 1, triangle ABC is equal to def 2, be = CF
Proof: ∵ ab ∥ ed, AC ∥ FD
∴∠B=∠E
∴∠ACB=∠EFD
∵FB=CE
∴FB+CF=CE+CF
That is BC = EF
∴△ABC≌△DEF(ASA)
∴AB=ED,AC=DF
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