如图5,在△ABC中,∠C=90°点M是斜边AB的中点,将一个直角的顶点置于M,角的两边分别与AC BC交于D E,过点D E分别作DG⊥AB EH⊥AB G H为垂足,求证(1)DG\MH=GM\EH(已证出)(2)AG=MH
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![如图5,在△ABC中,∠C=90°点M是斜边AB的中点,将一个直角的顶点置于M,角的两边分别与AC BC交于D E,过点D E分别作DG⊥AB EH⊥AB G H为垂足,求证(1)DG\MH=GM\EH(已证出)(2)AG=MH](/uploads/image/z/14505523-43-3.jpg?t=%E5%A6%82%E5%9B%BE5%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0C%3D90%C2%B0%E7%82%B9M%E6%98%AF%E6%96%9C%E8%BE%B9AB%E7%9A%84%E4%B8%AD%E7%82%B9%2C%E5%B0%86%E4%B8%80%E4%B8%AA%E7%9B%B4%E8%A7%92%E7%9A%84%E9%A1%B6%E7%82%B9%E7%BD%AE%E4%BA%8EM%2C%E8%A7%92%E7%9A%84%E4%B8%A4%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%8EAC+BC%E4%BA%A4%E4%BA%8ED+E%2C%E8%BF%87%E7%82%B9D+E%E5%88%86%E5%88%AB%E4%BD%9CDG%E2%8A%A5AB+EH%E2%8A%A5AB+G+H%E4%B8%BA%E5%9E%82%E8%B6%B3%2C%E6%B1%82%E8%AF%81%281%29DG%5CMH%3DGM%5CEH%28%E5%B7%B2%E8%AF%81%E5%87%BA%29%282%29AG%3DMH)
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