如图,BD是△ABC的中线,CE⊥BD于点E,AF⊥BD交BD的延长线于点F. (1)求证:DE=DF; (1)求证:BE+BF=2BD; (2)连AE、CF,求证:AE∥CF.
问题描述:
如图,BD是△ABC的中线,CE⊥BD于点E,AF⊥BD交BD的延长线于点F.
(1)求证:DE=DF;
(1)求证:BE+BF=2BD;
(2)连AE、CF,求证:AE∥CF.
答
(1)证明:∵BD是△ABC的中线,∴AD=CD,∵CE⊥BD,AF⊥BD,∴∠CED=∠AFD=90°,在△CDE和△ADF中,∠CED=∠AFD=90°∠ADF=∠CDEAD=CD,∴△CDE≌△ADF(AAS),∴DE=DF;(2)∵BE=BD-DE,BF=BD+DF,∴BE+BF=...