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
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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
==>∠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
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