已知△ABC中,M是BC中点,AE是角BAC的平分线,过B作BD垂直AE,垂足为D,交AC于B’,AM与BD相交于F,求证:EF平行AB
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![已知△ABC中,M是BC中点,AE是角BAC的平分线,过B作BD垂直AE,垂足为D,交AC于B’,AM与BD相交于F,求证:EF平行AB](/uploads/image/z/5447142-54-2.jpg?t=%E5%B7%B2%E7%9F%A5%E2%96%B3ABC%E4%B8%AD%2CM%E6%98%AFBC%E4%B8%AD%E7%82%B9%2CAE%E6%98%AF%E8%A7%92BAC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E8%BF%87B%E4%BD%9CBD%E5%9E%82%E7%9B%B4AE%2C%E5%9E%82%E8%B6%B3%E4%B8%BAD%2C%E4%BA%A4AC%E4%BA%8EB%E2%80%99%2CAM%E4%B8%8EBD%E7%9B%B8%E4%BA%A4%E4%BA%8EF%2C%E6%B1%82%E8%AF%81%3AEF%E5%B9%B3%E8%A1%8CAB)
已知△ABC中,M是BC中点,AE是角BAC的平分线,过B作BD垂直AE,垂足为D,交AC于B’,AM与BD相交于F,求证:EF平行AB
已知△ABC中,M是BC中点,AE是角BAC的平分线,过B作BD垂直AE,垂足为D,交AC于B’,AM与BD相交于F,求证:EF平行AB
已知△ABC中,M是BC中点,AE是角BAC的平分线,过B作BD垂直AE,垂足为D,交AC于B’,AM与BD相交于F,求证:EF平行AB
因为,∠ABD = 90°-∠DAB = 90°-∠DAB' = ∠AB'D ,
所以,AB = AB' ,
连接EB',
在等腰△ABB'中,顶角平分线AD是底边BB'的垂直平分线,即有:EB = EB' ,
可得:∠EBB' = ∠EB'B ;
在等腰△EBB'中,底边BB'上的高平分顶角∠BEB' ,即有:∠BED = ∠B'ED ;
连接DM,
因为,BD = B'D ,BM = CM ,
所以,DM是△BB'C的中位线,
可得:DM∥AC ,B'C = 2DM ,
则有:AC = AB'+B'C = AB'+2DM ;
已知,AE是∠BAC的平分线,可得:AB/AC = BE/CE ;
BF = BD+DF = B'D+DF = B'F+DF+DF = B'F+2DF ;
CE = CM+ME = BM+ME = BE+ME+ME = BM+2ME ;
因为,DM∥AC ,
所以,B'F/DF = AB'/DM ,
可得:B'F/(B'F+2DF) = AB'/(AB'+2DM) ,
则有:B'F/BF = AB'/AC = AB/AC = BE/CE ,
可得:B‘F/(B'F+BF) = BE/(BE+CE) ,
则有:B'F/BB' = BE/BC = B'E/BC ,
即有:B'F/B'E = BB'/BC ;
因为,在△B'EF和△BCB'中,∠EB'F = ∠CBB' , B'F/B'E = BB'/BC ,
所以,△B'EF ∽ △BCB' ,
可得:∠B'EF = ∠BCB' ;
因为,∠AEF = ∠AEB'-∠B'EF = ∠AEB-∠BCB' = ∠EAC = ∠EAB ,
所以,EF∥AB .