Question

如图所示,已知BD是∠ABC的平分线,AB=BC,点P在BD上,PM⊥AD,PN⊥CD,点M,N分别为垂足,且PM=PN

Answer 1

题目中应该没有AB=BC的吧?

证明：因为PM⊥AD PN⊥DC PM=PN PD=PD

∴△PMD≅△PND(HL)

∴∠PDM=∠PDN

又∠PDN=∠DBC+∠DCB

∠PDM=∠DBA+∠DAB

又因为∠DBA=∠DBC

∴∠DAB=∠DCB

又 BD=BD

∴△DBA≅△DBC

∴AB=BC