如图1,在正方形ABCD中,对角线AC与BD相交于点E,AF平分∠BAC,交BD于点F. (1)求证:EF+ AC=AB; (2)如图1,在正方形ABCD中,对角线AC与BD相交于点E,AF平分∠BAC,交BD于点F.(1)求证:EF+ AC=AB;(2)
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![如图1,在正方形ABCD中,对角线AC与BD相交于点E,AF平分∠BAC,交BD于点F. (1)求证:EF+ AC=AB; (2)如图1,在正方形ABCD中,对角线AC与BD相交于点E,AF平分∠BAC,交BD于点F.(1)求证:EF+ AC=AB;(2)](/uploads/image/z/14781610-10-0.jpg?t=%E5%A6%82%E5%9B%BE1%2C%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2C%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E4%B8%8EBD%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9E%2CAF%E5%B9%B3%E5%88%86%E2%88%A0BAC%2C%E4%BA%A4BD%E4%BA%8E%E7%82%B9F%EF%BC%8E+%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9AEF%2B+AC%3DAB%EF%BC%9B+%EF%BC%882%EF%BC%89%E5%A6%82%E5%9B%BE1%2C%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2C%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E4%B8%8EBD%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9E%2CAF%E5%B9%B3%E5%88%86%E2%88%A0BAC%2C%E4%BA%A4BD%E4%BA%8E%E7%82%B9F%EF%BC%8E%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9AEF%2B+AC%3DAB%EF%BC%9B%EF%BC%882%EF%BC%89)
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