已知三角形ABC,AB=AC,H是BC中点,HE垂直于AC于E,O为HE中点,求证：AO垂直于BE
2个回答

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.

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