题目:在△ABC中,∠ACB=90度(AC<BC),AD为BC边上的中线,E是AD的中点,CE延长线交AB于F,FG‖AC交AD于G.求证:FB=2AF.答案:证明:延长AD到H,使DH=AD,连结CH,BH.∵AD平分BC,即BD=CD而DH=AD∴四边形ACHB为平行四边形∴AB‖CH,AB=CH∴∠EAF=∠CHE,∠AFE=∠HCE∴△AFE∽△HCE∴AF/CH=AE/EH∵E为AD中点∴AE=DE=1/2AD而DH=AD∴EH=CE+DH=1/2AD+AD=3/2AD∴AE/EH=1/3∴AF/CH=1/3,即AF=1/3CH∵CH=AB∴AF=1/3AB∵BF+AF=AB∴BF=AB-AF=AB-1/3AB=2/3AB∴BF=2AF