摘要:证明:∵四边形ABCD是正方形.AE⊥BF ∴∠DAE+∠AED = 90°.∠DAE+∠AFB = 90° ∴∠AED = ∠AFB 又∵AD = AB.∠BAD = ∠D . ∴△AED≌△ABF ∴AE = BF

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