点P是△ABC中位线MN上任意一点,BP,CP的延长线分别交对边AC,AB于点D,E.求证:AD:DC+AE:EB=1
问题描述:
点P是△ABC中位线MN上任意一点,BP,CP的延长线分别交对边AC,AB于点D,E.求证:AD:DC+AE:EB=1
答
延长AP交BC于F,再过F作FG∥CE交AB于G、作FH∥BD交AC于H.
∵MN是△ABC中过AB、AC的中位线,∴MN∥BC,∴MP∥BF,∴AP=PF.
∵FG∥CE、AP=PF,∴AE=EG. ∵FH∥BD、AP=PF,∴AD=DH.
由FG∥CE,得:EG/EB=CF/BC,∴AE/EB=CF/BC.
由FH∥BD,得:DH/DC=BF/BC,∴AD/DC=BF/BC.
由AE/EB=CF/BC、AD/DC=BF/BC,得:AE/EB+AD/DC=(CF+BF)/BC=1.
即:AE∶EB+AD∶DC=1.