已知,如图c为线段ab上一点,分别以ac和bc为边做等边三角形acd和等边三角形bce,连接ae、bd,交cd于g,bd交ce于h,连fc、gh.(1)求角dfc的度数(2)求证fa=fd+fc
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![已知,如图c为线段ab上一点,分别以ac和bc为边做等边三角形acd和等边三角形bce,连接ae、bd,交cd于g,bd交ce于h,连fc、gh.(1)求角dfc的度数(2)求证fa=fd+fc](/uploads/image/z/8832028-4-8.jpg?t=%E5%B7%B2%E7%9F%A5%2C%E5%A6%82%E5%9B%BEc%E4%B8%BA%E7%BA%BF%E6%AE%B5ab%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E5%88%86%E5%88%AB%E4%BB%A5ac%E5%92%8Cbc%E4%B8%BA%E8%BE%B9%E5%81%9A%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2acd%E5%92%8C%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2bce%2C%E8%BF%9E%E6%8E%A5ae%E3%80%81bd%2C%E4%BA%A4cd%E4%BA%8Eg%2Cbd%E4%BA%A4ce%E4%BA%8Eh%2C%E8%BF%9Efc%E3%80%81gh.%EF%BC%881%EF%BC%89%E6%B1%82%E8%A7%92dfc%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81fa%3Dfd%2Bfc)
已知,如图c为线段ab上一点,分别以ac和bc为边做等边三角形acd和等边三角形bce,连接ae、bd,交cd于g,bd交ce于h,连fc、gh.(1)求角dfc的度数(2)求证fa=fd+fc
已知,如图c为线段ab上一点,分别以ac和bc为边做等边三角形acd和等边三角形bce,连接ae、bd,交cd于g,bd交ce于h,连fc、gh.
(1)求角dfc的度数
(2)求证fa=fd+fc
已知,如图c为线段ab上一点,分别以ac和bc为边做等边三角形acd和等边三角形bce,连接ae、bd,交cd于g,bd交ce于h,连fc、gh.(1)求角dfc的度数(2)求证fa=fd+fc
看不出什么等腰三角形,不过直角是有的,这题.不需要去证明.
思路复杂,我一段一段的画图给你看
因为 AC = AD, BC = CE, ∠ACD + ∠CDE = ∠ECB + ∠CDE,所以 △ACE ≌ △DCB
=> ∠1 = ∠1' & ∠2 = ∠2' & AE = BD (有些题目会出 AE = BD)
△GDF & △ACG,因为 ∠2 = ∠2' & 对顶角 ∠DGF = ∠AGC,所以 ∠3 = 60° & 补角∠GFH = 120°
因为 ∠2 = ∠2',∠ACD = ∠DCE = 60°,AC = AD,所以△ACG ≌ △DCH
=> CG = CH => △CGH为等边△ => ∠4 = 60°
四边形CHFG,因为∠GCH + ∠GFH = 60° + 120°,所以 四点共圆
=>
同弦圆周角 ∠GFC = ∠4 = 60°
∠DFC = ∠3 + ∠4 = 60° + 60° = 120°
AE上取一点I,使 FI = FC
因为FI = FC,∠5 = 60°,所以△IFC为等边△ => ∠ICF = 60°
∠7 = 60° - ∠ICG = ∠ICF - ∠ICG = ∠6
因为AC = CD,∠7 = ∠6,∠2 = ∠2',所以△AIC ≌ △DFC
=> AI = DF
AF = AI + IF = DF + FC
=> AF = DF + FC