已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(答好有追问)已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(1)CE=CF (2)E
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![已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(答好有追问)已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(1)CE=CF (2)E](/uploads/image/z/5423664-48-4.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%A6%82%E5%9B%BE1%2C%E7%82%B9C%E4%B8%BA%E7%BA%BF%E6%AE%B5AB%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E2%96%B3ACM%2C%E2%96%B3CBN%E9%83%BD%E6%98%AF%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2%2CAN%E4%BA%A4MC%E4%BA%8E%E7%82%B9E%2CBM%E4%BA%A4CN%E4%BA%8E%E7%82%B9F%EF%BC%8E%28%E7%AD%94%E5%A5%BD%E6%9C%89%E8%BF%BD%E9%97%AE%EF%BC%89%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%A6%82%E5%9B%BE1%2C%E7%82%B9C%E4%B8%BA%E7%BA%BF%E6%AE%B5AB%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E2%96%B3ACM%2C%E2%96%B3CBN%E9%83%BD%E6%98%AF%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2%2CAN%E4%BA%A4MC%E4%BA%8E%E7%82%B9E%2CBM%E4%BA%A4CN%E4%BA%8E%E7%82%B9F%EF%BC%8E%281%29CE%3DCF+%282%29E)
已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(答好有追问)已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(1)CE=CF (2)E
已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(答好有追问)
已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.
(1)CE=CF
(2)EF∥AB
要过程
答好有追加悬赏
已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(答好有追问)已知:如图1,点C为线段AB上一点,△ACM,△CBN都是等边三角形,AN交MC于点E,BM交CN于点F.(1)CE=CF (2)E
∠BAM=∠BCN=60°,∠ACM=∠ABN=60°
=>AM//CN,CM//BN
∠AEM=∠CEN,∠CFM=∠BFN
=>△AEM∽△CEN,△CFM∽△BFN
=>ME:EC=AM:CN,MF:FB=CM:BN
AM=CM,CN=BN
=>AM:CN=CM:BN
=>ME:EC=MF:FB
=>EF//CB (2)
=>∠CFE=∠BCF,∠CEF=∠ACE
=>∠CEF=∠CFE
=>CE=CF (1)
图呢?
∠BAM=∠BCN=60°,∠ACM=∠ABN=60°
=>AM//CN,CM//BN
∠AEM=∠CEN,∠CFM=∠BFN
=>△AEM∽△CEN,△CFM∽△BFN
=>ME:EC=AM:CN,MF:FB=CM:BN
AM=CM,CN=BN
=>AM:CN=CM:BN
=>ME:EC=MF:FB
=>EF//CB (2)
=>∠CFE=∠BCF,∠CEF=∠ACE
=>∠CEF=∠CFE
=>CE=CF (1)