已知如图,△ABC中,以AB为直径作⊙O,交BC于D,交AC于E.过D点作⊙O的切线FG交AC于F,交AB的延长线于G,连接AD.若AB:BG=3:1,FG⊥AC. (1)求证:AD平分∠CAB; (2)若GD=4,求BD; (3)求AE

问题描述:

已知如图,△ABC中,以AB为直径作⊙O,交BC于D,交AC于E.过D点作⊙O的切线FG交AC于F,交AB的延长线于G,连接AD.若AB:BG=3:1,FG⊥AC.
(1)求证:AD平分∠CAB;
(2)若GD=4,求BD;
(3)求AE:EF:FC.

(1)证明:∵GF是切线,∴OD⊥GF∴∠ODF=90°即∠ODA+∠ADF=90°∵GF⊥AC∴∠AFG=90°即∠ADF+∠DAC=90°∴∠ODA=∠DAC∵∠ODA=∠OAD∴∠DAC=∠ODA∴AD平分∠CAB;(2)∵GD是⊙O的切线,由切割线定理得:GD2=GB•G...