1.已知ABCD是正方形,对角线AC,BD相交于O,四边形AEFC是菱形,EH⊥AC,垂足为H,求证:EH=1/2FC2.以Rt△ABC的两直角边AB,AC向外作正方形ABDM,ACEN,由∠BAM,∠CAN的对角的顶点D.E分别向斜边所在直线作垂线DF,EG,垂
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![1.已知ABCD是正方形,对角线AC,BD相交于O,四边形AEFC是菱形,EH⊥AC,垂足为H,求证:EH=1/2FC2.以Rt△ABC的两直角边AB,AC向外作正方形ABDM,ACEN,由∠BAM,∠CAN的对角的顶点D.E分别向斜边所在直线作垂线DF,EG,垂](/uploads/image/z/2821733-53-3.jpg?t=1.%E5%B7%B2%E7%9F%A5ABCD%E6%98%AF%E6%AD%A3%E6%96%B9%E5%BD%A2%2C%E5%AF%B9%E8%A7%92%E7%BA%BFAC%2CBD%E7%9B%B8%E4%BA%A4%E4%BA%8EO%2C%E5%9B%9B%E8%BE%B9%E5%BD%A2AEFC%E6%98%AF%E8%8F%B1%E5%BD%A2%2CEH%E2%8A%A5AC%2C%E5%9E%82%E8%B6%B3%E4%B8%BAH%2C%E6%B1%82%E8%AF%81%3AEH%3D1%2F2FC2.%E4%BB%A5Rt%E2%96%B3ABC%E7%9A%84%E4%B8%A4%E7%9B%B4%E8%A7%92%E8%BE%B9AB%2CAC%E5%90%91%E5%A4%96%E4%BD%9C%E6%AD%A3%E6%96%B9%E5%BD%A2ABDM%2CACEN%2C%E7%94%B1%E2%88%A0BAM%2C%E2%88%A0CAN%E7%9A%84%E5%AF%B9%E8%A7%92%E7%9A%84%E9%A1%B6%E7%82%B9D.E%E5%88%86%E5%88%AB%E5%90%91%E6%96%9C%E8%BE%B9%E6%89%80%E5%9C%A8%E7%9B%B4%E7%BA%BF%E4%BD%9C%E5%9E%82%E7%BA%BFDF%2CEG%2C%E5%9E%82)
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