Let D, E points onto the sides BC, AB of the equilateral triangle ABC, and F onto (BA so that CD=BE=AF, and K midpoint of DE. Find <CKF.
Proof:
Construct the equilateral triangle BEG, outside ABC, DCEG is a parallelogram, see that triangles FEG and CAF are congruent, s.a.s, thus FG=FC, and the median FK of the isosceles triangle CFG is perpendicular bisector as well.
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