摘要： 如图.在△ABC中.∠ACB=90°.BC的垂直平分线DE交BC于D.交AB于E.F在DE上.且AF=CE=AE． ⑴说明四边形ACEF是平行四边形, ⑵当∠B满足什么条件时.四边形ACEF是菱形.并说明理由． [答案](1)证明:由题意知∠FDC =∠DCA = 90°．∴EF∥CA ∴∠AEF =∠EAC ∵AF = CE = AE ∴∠F =∠AEF =∠EAC =∠ECA 又∵AE = EA ∴△AEC≌△EAF.∴EF = CA.∴四边形ACEF是平行四边形 ． (2)当∠B=30°时.四边形ACEF是菱形 ． 理由是:∵∠B=30°.∠ACB=90°.∴AC=.∵DE垂直平分BC.∴ BE=CE 又∵AE=CE.∴CE=.∴AC=CE.∴四边形ACEF是菱形．

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