如图,已知矩形ABCD中,E是CD上一点,AD∶AE=1∶2,CE∶ED=1∶3 (1)求证:AE⊥BE (2)F是AB中点,DF交AE于G,若CE =G根号3,求S△DFA.
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![如图,已知矩形ABCD中,E是CD上一点,AD∶AE=1∶2,CE∶ED=1∶3 (1)求证:AE⊥BE (2)F是AB中点,DF交AE于G,若CE =G根号3,求S△DFA.](/uploads/image/z/3579199-7-9.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%E7%9F%A9%E5%BD%A2ABCD%E4%B8%AD%2CE%E6%98%AFCD%E4%B8%8A%E4%B8%80%E7%82%B9%2CAD%E2%88%B6AE%EF%BC%9D1%E2%88%B62%2CCE%E2%88%B6ED%3D1%E2%88%B63+%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%3AAE%E2%8A%A5BE+%EF%BC%882%EF%BC%89F%E6%98%AFAB%E4%B8%AD%E7%82%B9%2CDF%E4%BA%A4AE%E4%BA%8EG%2C%E8%8B%A5CE+%3DG%E6%A0%B9%E5%8F%B73%2C%E6%B1%82S%E2%96%B3DFA.)
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