点C是线段BD上一点(与B ,D不重复),AB=AC,DE⊥直线AC,垂足E,求证;BC×CD=2AC×CE
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![点C是线段BD上一点(与B ,D不重复),AB=AC,DE⊥直线AC,垂足E,求证;BC×CD=2AC×CE](/uploads/image/z/12550998-30-8.jpg?t=%E7%82%B9C%E6%98%AF%E7%BA%BF%E6%AE%B5BD%E4%B8%8A%E4%B8%80%E7%82%B9%28%E4%B8%8EB+%2CD%E4%B8%8D%E9%87%8D%E5%A4%8D%29%2CAB%3DAC%2CDE%E2%8A%A5%E7%9B%B4%E7%BA%BFAC%2C%E5%9E%82%E8%B6%B3E%2C%E6%B1%82%E8%AF%81%3BBC%C3%97CD%3D2AC%C3%97CE)
点C是线段BD上一点(与B ,D不重复),AB=AC,DE⊥直线AC,垂足E,求证;BC×CD=2AC×CE
点C是线段BD上一点(与B ,D不重复),AB=AC,DE⊥直线AC,垂足E,求证;BC×CD=2AC×CE
点C是线段BD上一点(与B ,D不重复),AB=AC,DE⊥直线AC,垂足E,求证;BC×CD=2AC×CE
过A点做AF⊥BD
∵AB = AC, AF⊥BD
∴ BF= CF,即CF=1/2 BC
又∵∠ACF= ∠DCE
∠AFC= ∠CED=90
△AFC∽△DCE
∴AC/CD=CF/CE,即AC*CE=CF*CD
又∵CF=1/2 BC
∴2AC*CE=BC*CD
一楼做得对