如图,在△ABC中,AB>AC,边AB上取一点D,边AC上取一点E,使AD=AE,直线DE和BC的延长线交于点P.求证:BP:CP=BD:CE.

问题描述:

如图,在△ABC中,AB>AC,边AB上取一点D,边AC上取一点E,使AD=AE,直线DE和BC的延长线交于点P.求证:BP:CP=BD:CE.

证明:如图,过点B作BF∥AC交PD延长线于点F.则△PCE∽△PBF,

BF
CE
=
BP
CP

∵BF∥AC,
∴∠1=∠2.
又∵AD=AE,
∴∠2=∠4,
∠1=∠3=∠4,
∴BF=BD.
BF
CE
=
BD
CE

∴BP:CP=BD:CE.