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过B作BD⊥AC交AC于D.令AO、BE相交于F.
∵AB=AC、BH=CH,∴∠AHC=90°.
∵∠AHC=∠BDC=90°,∴∠EAH=∠DBC(同是∠C的余角).
∵∠EAH=∠DBC、∠AEH=∠BDC=90°,∴△EAH∽△DBC,∴AE/BD=EH/DC.······①
∵BD⊥AC、HE⊥AC,∴BD∥HE,又BH=CH,∴DE=DC/2,而EO=EH/2,
∴EH/DC=(EH/2)/(DC/2)=EO/DE.······②
由①、②,得:AE/BD=EO/DE,又∠AEO=∠BDE=90°,∴△AEO∽△BDE,
∴∠DAF=∠DBF,∴A、B、F、D共圆,∴∠AFB=∠ADB=90°,∴AO⊥BE.
1年前
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