如图,在△ABC中,AD为∠A的平分线.EF是AD的垂直平分线,且交AB于E,交BC得延长线于F.求证DF²=CF×BF

连接AF∵EF是AD的垂直平分线∴AF=DF,∠FAD=∠FDA∵AD为∠A的平分线∴∠BAD =∠CAD∵∠FDA=∠ABD+∠BAD∠FAD=∠CAF +∠CAD∴∠CAF =∠ABD∵∠CAF+∠ACF+∠AFC=180°∠ABD +∠BAF+∠AFC=180°∴∠ACF =∠BAF∴△BAF相...