如图,AB=AC,AD=AE,BE和CD相交于点o,AO的延长线叫BC与点F,求证:BF=FC图放不上来
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![如图,AB=AC,AD=AE,BE和CD相交于点o,AO的延长线叫BC与点F,求证:BF=FC图放不上来](/uploads/image/z/10158842-2-2.jpg?t=%E5%A6%82%E5%9B%BE%2CAB%3DAC%2CAD%3DAE%2CBE%E5%92%8CCD%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9o%2CAO%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E5%8F%ABBC%E4%B8%8E%E7%82%B9F%2C%E6%B1%82%E8%AF%81%3ABF%3DFC%E5%9B%BE%E6%94%BE%E4%B8%8D%E4%B8%8A%E6%9D%A5)
如图,AB=AC,AD=AE,BE和CD相交于点o,AO的延长线叫BC与点F,求证:BF=FC图放不上来
如图,AB=AC,AD=AE,BE和CD相交于点o,AO的延长线叫BC与点F,求证:BF=FC
图放不上来
如图,AB=AC,AD=AE,BE和CD相交于点o,AO的延长线叫BC与点F,求证:BF=FC图放不上来
证明:
∵AB=AC,∠BAE=∠CAD,AE=AD
∴⊿ABE≌⊿ACD(SAS)
∴∠ABE=∠ACD
∵AB=AC
∴∠ABC=∠ACB
∴∠ABC-∠ABE=∠ACB-∠ACD
即∠OBF=∠OCF
∴OB=OC
又∵AB=AC,AO=AO
∴⊿ABO≌⊿ACO(SSS)
∴∠BAO=∠CAO
即AF平分∠BAC
∴BF=CF【三线合一,等腰三角形顶角平分线也是底边中线】
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