AB=AC,∠A=120°
==>∠B = ∠C =30°
EF是AB的垂直平分线
==>BF = AF,∠B = ∠FAB = 30°
==>∠FAC = 90°
==>AF/CF = sin∠C = 1/2
==>CF = 2AF = 2BF
在三角形ABC中,AB=AC,角A=120度,AB的垂直平分线交AB于点E,交BC于点F,求证CF=2BF
在三角形ABC中,AB=AC,角A=120度,AB的垂直平分线交AB于点E,交BC于点F,求证CF=2BF
数学人气：251 ℃时间：2019-08-19 12:06:57
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