一道初二几何题,如图:Rt△ABC中,∠C=90°,CD⊥AB,BE平分∠ABC交AC于E,交CD于F,EH⊥CD于H,则下列结论:①AC^2+BD^2=BC^2+AD^2 ②BD+EH/BC为定值 ③若F为BE中点,则AD=3BD,其中正确的结论有:点击图片可以看到
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![一道初二几何题,如图:Rt△ABC中,∠C=90°,CD⊥AB,BE平分∠ABC交AC于E,交CD于F,EH⊥CD于H,则下列结论:①AC^2+BD^2=BC^2+AD^2 ②BD+EH/BC为定值 ③若F为BE中点,则AD=3BD,其中正确的结论有:点击图片可以看到](/uploads/image/z/13770854-62-4.jpg?t=%E4%B8%80%E9%81%93%E5%88%9D%E4%BA%8C%E5%87%A0%E4%BD%95%E9%A2%98%2C%E5%A6%82%E5%9B%BE%EF%BC%9ARt%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0C%3D90%C2%B0%2CCD%E2%8A%A5AB%2CBE%E5%B9%B3%E5%88%86%E2%88%A0ABC%E4%BA%A4AC%E4%BA%8EE%2C%E4%BA%A4CD%E4%BA%8EF%2CEH%E2%8A%A5CD%E4%BA%8EH%2C%E5%88%99%E4%B8%8B%E5%88%97%E7%BB%93%E8%AE%BA%EF%BC%9A%E2%91%A0AC%5E2%2BBD%5E2%3DBC%5E2%2BAD%5E2+%E2%91%A1BD%2BEH%2FBC%E4%B8%BA%E5%AE%9A%E5%80%BC+%E2%91%A2%E8%8B%A5F%E4%B8%BABE%E4%B8%AD%E7%82%B9%2C%E5%88%99AD%3D3BD%2C%E5%85%B6%E4%B8%AD%E6%AD%A3%E7%A1%AE%E7%9A%84%E7%BB%93%E8%AE%BA%E6%9C%89%EF%BC%9A%E7%82%B9%E5%87%BB%E5%9B%BE%E7%89%87%E5%8F%AF%E4%BB%A5%E7%9C%8B%E5%88%B0)
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