如图,△ABC中,∠ACB=90°,CD⊥AB于D,AO平分∠BAC,交CD于O,E为AB上一点,且AE=AC,求证:OE∥BC
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![如图,△ABC中,∠ACB=90°,CD⊥AB于D,AO平分∠BAC,交CD于O,E为AB上一点,且AE=AC,求证:OE∥BC](/uploads/image/z/5901926-14-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0ACB%3D90%C2%B0%2CCD%E2%8A%A5AB%E4%BA%8ED%2CAO%E5%B9%B3%E5%88%86%E2%88%A0BAC%2C%E4%BA%A4CD%E4%BA%8EO%2CE%E4%B8%BAAB%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E4%B8%94AE%3DAC%2C%E6%B1%82%E8%AF%81%3AOE%E2%88%A5BC)
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