2011-9-25 如图,BD是∠ABC的角平分线,AD=CD,求证∠DAB+∠BCD=180°

问题描述:

2011-9-25 如图,BD是∠ABC的角平分线,AD=CD,求证∠DAB+∠BCD=180°

过D点作AB垂线DE,过D点作BC垂线DF
∵BD是∠ABC的平分线,且AD=CD

∴△AED≌△CFD
∴∠EAD=∠FCD
∠EAD+∠DAB=180°
∠EAD=∠FCD=∠BCD

∴∠DAB+∠BCD=180°