Question

如图1-2-5,在△ABC中,AB＞AC,D在AB上,E在AC上且AD=AE,直线DE和BC的延长线交于点P.

求证:BP∶CP=BD∶CE.

图1-2-5

Answer 1

证明:过C作CF∥AB,交DP于F,则BP∶CP=BD∶CF,

∠EFC=∠ADE.

∵AD=AE,

∴∠ADE=∠AED.

∴∠AED=∠CFE.

∵∠AED=∠CEF,

∴∠CEF=∠CFE.

∴CE=CF.

∴BP∶CP=BD∶CE.