如图,ΔABC中,∠ACB=90°,CD垂直AB于点D,AE平分∠CAB,交CD于F,G为AB上一点,且AC=AG.试说明FG∥BC
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![如图,ΔABC中,∠ACB=90°,CD垂直AB于点D,AE平分∠CAB,交CD于F,G为AB上一点,且AC=AG.试说明FG∥BC](/uploads/image/z/6960103-7-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%CE%94ABC%E4%B8%AD%2C%E2%88%A0ACB%3D90%C2%B0%2CCD%E5%9E%82%E7%9B%B4AB%E4%BA%8E%E7%82%B9D%2CAE%E5%B9%B3%E5%88%86%E2%88%A0CAB%2C%E4%BA%A4CD%E4%BA%8EF%2CG%E4%B8%BAAB%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E4%B8%94AC%3DAG.%E8%AF%95%E8%AF%B4%E6%98%8EFG%E2%88%A5BC)
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