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D. E, f are the points on each side of △ ABC, and DF / / AB, de / / BC, we know ∠ B = 60 °, find the degree of ∠ EDF, and explain the reason
Because DF is parallel to AB and De is parallel to BC, edfb is a parallelogram. Angle B = 60. That angle EDF = 60
As shown in the figure, it is known that point D is a point on the BC side of the regular triangle ABC, ∠ EDF = 120 °, ED, DF intersect AB and AC respectively at points e, F, CD = dB
Be + CF = BD
Take a P on AB and make Δ PBD an equilateral triangle. Then ∠ BPD = 60 ° = ∠ C, PD = BD, ∵ BC = CD, ∵ PD = CD = Pb, ∵ EDF = 120 °, ∵ BAC = 60 °, ∵ AED + ∠ AFD = 180 ° (the sum of the inner angles of the quadrilateral is 360 °),
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