三角形ABC,已知角A是锐角,在AB上取一点D,AC上取一点E,连接CD,BE,使角DCB=角EBC=1/2角A,求BD与CE数量关
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![三角形ABC,已知角A是锐角,在AB上取一点D,AC上取一点E,连接CD,BE,使角DCB=角EBC=1/2角A,求BD与CE数量关](/uploads/image/z/3781644-60-4.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2ABC%2C%E5%B7%B2%E7%9F%A5%E8%A7%92A%E6%98%AF%E9%94%90%E8%A7%92%2C%E5%9C%A8AB%E4%B8%8A%E5%8F%96%E4%B8%80%E7%82%B9D%2CAC%E4%B8%8A%E5%8F%96%E4%B8%80%E7%82%B9E%2C%E8%BF%9E%E6%8E%A5CD%2CBE%2C%E4%BD%BF%E8%A7%92DCB%3D%E8%A7%92EBC%3D1%2F2%E8%A7%92A%2C%E6%B1%82BD%E4%B8%8ECE%E6%95%B0%E9%87%8F%E5%85%B3)
三角形ABC,已知角A是锐角,在AB上取一点D,AC上取一点E,连接CD,BE,使角DCB=角EBC=1/2角A,求BD与CE数量关
三角形ABC,已知角A是锐角,在AB上取一点D,AC上取一点E,连接CD,BE,使角DCB=角EBC=1/2角A,求BD与CE数量关
三角形ABC,已知角A是锐角,在AB上取一点D,AC上取一点E,连接CD,BE,使角DCB=角EBC=1/2角A,求BD与CE数量关
在CD上或CD的延长线上取一点F,使得:CF = BE .
在△CBF和△BCE中,CF = BD ,∠BCF = ∠CBE ,BC为公共边,
所以,△CBF ≌ △BCE ,可得:BF = CE ,∠CBF = ∠BCE ,∠CFB = ∠BEC .
① 若点F在CD上,则有:
∠BFD = ∠CBF+∠DCB = ∠BCE+∠DCB = ∠ACD+2∠DCB = ∠ACD+∠A = ∠BDF ;
可得:BD = BF = CE ;
② 若点F在CD的延长线上,则有:
∠BFD = ∠BEC = ∠ABE+∠A = ∠ABE+∠EBC+∠DCB = ∠ABC+∠DCB = ∠ADC = ∠BDF ;
可得:BD = BF = CE ;
③ 若点F和点D重合,
可得:BD = BF = CE ;
综上可得:BD = CE .