如图,在Rt△ABC中,∠BAC=90°,AD⊥BC于点D,角平分线BE交AD于点F,过点F做FG∥BC,交AC于点G,求证:AE=CG.
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![如图,在Rt△ABC中,∠BAC=90°,AD⊥BC于点D,角平分线BE交AD于点F,过点F做FG∥BC,交AC于点G,求证:AE=CG.](/uploads/image/z/5199860-20-0.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8Rt%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0BAC%3D90%C2%B0%2CAD%E2%8A%A5BC%E4%BA%8E%E7%82%B9D%2C%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BFBE%E4%BA%A4AD%E4%BA%8E%E7%82%B9F%2C%E8%BF%87%E7%82%B9F%E5%81%9AFG%E2%88%A5BC%2C%E4%BA%A4AC%E4%BA%8E%E7%82%B9G%2C%E6%B1%82%E8%AF%81%3AAE%3DCG.)
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