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梯形ABCD中,AB//CD,以AD、AC为边做平行四边形ACED,延长DC交EB于F,求证：EF=FB
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梯形ABCD中,AB//CD,以AD、AC为边做平行四边形ACED,延长DC交EB于F,求证：EF=FB
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答案和解析
过A做AG//BE,交CD于G
∴∠DGA = ∠DFB
∵AB//CD
∴四边形AGFB是平行四边形
∴AG = BF
∵ADEC是平行四边形
∴AD = CE
∠ADG = ∠ECF
∵∠DFB = ∠CFE
∴∠DGA = ∠CFE
∴△ADG≌△ECF
∴AG = FE
∵AG = FB
∴FE = FB
∴∠DGA = ∠DFB
∵AB//CD
∴四边形AGFB是平行四边形
∴AG = BF
∵ADEC是平行四边形
∴AD = CE
∠ADG = ∠ECF
∵∠DFB = ∠CFE
∴∠DGA = ∠CFE
∴△ADG≌△ECF
∴AG = FE
∵AG = FB
∴FE = FB
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