【2020 长宁金山一模】 23.(本题满分12 分,第(1)小题5 分,第(2)小题7 分) 如图,在ABC中,点D 、E 分别在边AB、BC 上, AE与CD 交于点F .若AE平分BAC,AEACAFAB. (1)求证:AECAFD; (2)若CDEG //,交边AC 的延长线于点G , 求证:BDFCCGCD. (长宁金山)23.(本题满分12 分,第(1)小题5 分,第(2)小题7 分) 证明:(1) AEACAFAB ∴AFAEACAB (1 分) AE平分BAC ∴CAFBAE (1 分) ∴ ABE∽ ACF (1 分) ∴ACFB (1 分) 又 BAEBAECCAFACFAFD, ∴AECAFD (1 分) (2) AECAFD,CFEAFD ∴AECCFE (1 分) ∴CEFC (1 分) CDEG // ∴CEGDCB GACF 又 BACF ∴GB (2 分) ∴ BCD∽ GEC (1 分) ∴CGBDCECD (1 分) ∴CGBDFCCD 即BDFCCGCD. (1 分) 第23 题图 G A C B E D F 【2020 杨浦一模】 23.(本题满分12 分,每小题各6 分) 如图,已知在ABC△中,AD 是ABC△的中线,DACB ,点E在边AD 上,CECD. (1)求证:ACBDABAD; (2)求证:22ACAE AD. (杨浦)23.证明:(1) CD=CE,∴∠CED=∠CDA. ··········································· (1 分) ∴∠AEC=∠BDA. ······················································································ (1 分) 又 ∠DAC=∠B,∴△ACE∽△BAD. ························································· (1 分) ∴ACCEABAD. ···························································································· (1 分) AD 是ABC△的中线,∴BDCD . ····················································· (1 分) CD=CE,∴BDCE.∴ACBDABAD. ·················...