如图,已知三角形ABC的内角平分线BD与外角平分线CD相交于点D.求证：角A=2角D

Question

如图,已知三角形ABC的内角平分线BD与外角平分线CD相交于点D.求证：角A=2角D

Answer 1

∠ACE=∠ABC+∠A         

∠DCE=∠D+∠DBE 

∠ABC =2∠DBE 

所以

2∠DCE=2∠D+2∠DBE

2∠D+2∠DBE=∠ABC+∠A

∠A=2∠D