如图,在△ABC中,AD是∠BAC的平分线,M是BC的中点,过点M作ME//AD交BA的延长线于点E,交AC于点F.求证:BE=CF=1/2(AB+AC)
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![如图,在△ABC中,AD是∠BAC的平分线,M是BC的中点,过点M作ME//AD交BA的延长线于点E,交AC于点F.求证:BE=CF=1/2(AB+AC)](/uploads/image/z/6646149-45-9.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAD%E6%98%AF%E2%88%A0BAC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2CM%E6%98%AFBC%E7%9A%84%E4%B8%AD%E7%82%B9%2C%E8%BF%87%E7%82%B9M%E4%BD%9CME%2F%2FAD%E4%BA%A4BA%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8E%E7%82%B9E%2C%E4%BA%A4AC%E4%BA%8E%E7%82%B9F.%E6%B1%82%E8%AF%81%EF%BC%9ABE%3DCF%3D1%2F2%EF%BC%88AB%2BAC%EF%BC%89)
如图,在△ABC中,AD是∠BAC的平分线,M是BC的中点,过点M作ME//AD交BA的延长线于点E,交AC于点F.求证:BE=CF=1/2(AB+AC)
如图,在△ABC中,AD是∠BAC的平分线,M是BC的中点,过点M作ME//AD交BA的延长线于点E,交AC于点F.
求证:BE=CF=1/2(AB+AC)
如图,在△ABC中,AD是∠BAC的平分线,M是BC的中点,过点M作ME//AD交BA的延长线于点E,交AC于点F.求证:BE=CF=1/2(AB+AC)
证明:
过B作BN//AC交EM延长线于N点
∴CF=BN ∠CFM=∠N
又AD//ME AD平分∠BAC,∠E=∠DAB
∴∠CFM=∠DAC=∠E
∴∠E=∠N
∴ BE=BN=CF
∵∠EFA=∠CFM
∴∠E=∠EFA
∴ AE=AF
∴ AB+AC=AB+(AF+FC)=AB+AE+FC=BE+FC
∴ BE=CF=1/2(AB+AC)
证明:在EM延长线上取点H,使MH=MF,过点C作CG∥ME交BE的延长线于点G,连接BF、CH∵AD平分∠BAC∴∠BAD=∠CAD∵MN∥AD∴∠AEF=∠BAD,AFE=∠CAD∴∠AEF=∠AFE∵M是BC的中点∴BM=CM∵MH=MF∴平行四边形FBHC∴BH∥AC,BH=CF∴∠BHE=∠AFE∴∠BHE=∠AEF∴BH=BE∴BE=CF∵CG∥AD∴∠G=∠AEF,∠ACG=∠AFE...
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证明:在EM延长线上取点H,使MH=MF,过点C作CG∥ME交BE的延长线于点G,连接BF、CH∵AD平分∠BAC∴∠BAD=∠CAD∵MN∥AD∴∠AEF=∠BAD,AFE=∠CAD∴∠AEF=∠AFE∵M是BC的中点∴BM=CM∵MH=MF∴平行四边形FBHC∴BH∥AC,BH=CF∴∠BHE=∠AFE∴∠BHE=∠AEF∴BH=BE∴BE=CF∵CG∥AD∴∠G=∠AEF,∠ACG=∠AFE∴∠G=∠ACG∴AG=AC∴BG=AB+AG=AB+AC∵MN∥AD,CG∥AD∴CG∥MN又∵M是BC的中点∴中位线ME∴BE=BG/2=(AB+BC)/2∴BE=CF
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