已知:如图,在正方形ABCD外取一点E,连接AE,BE,DE,DE交AB于F.(2)若AE⊥AG,BE⊥DE,点F为AB的中点,求证:FG-EF=BE
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![已知:如图,在正方形ABCD外取一点E,连接AE,BE,DE,DE交AB于F.(2)若AE⊥AG,BE⊥DE,点F为AB的中点,求证:FG-EF=BE](/uploads/image/z/5036604-60-4.jpg?t=%E5%B7%B2%E7%9F%A5%3A%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E5%A4%96%E5%8F%96%E4%B8%80%E7%82%B9E%2C%E8%BF%9E%E6%8E%A5AE%2CBE%2CDE%2CDE%E4%BA%A4AB%E4%BA%8EF.%282%29%E8%8B%A5AE%E2%8A%A5AG%2CBE%E2%8A%A5DE%2C%E7%82%B9F%E4%B8%BAAB%E7%9A%84%E4%B8%AD%E7%82%B9%2C%E6%B1%82%E8%AF%81%EF%BC%9AFG-EF%3DBE)
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