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°
答
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°