已知MN是三角形ABC的中位线,P在MN上,BP,CP交对边于D,E.求证AE:BE+AD:DC=1
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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.