Side CD of parallelogram ABCD is twice as large as side BC. Point F is the middle of the side of the CD. Prove that BF is the bisector of the angle ABC

BC = CD / 2 = CF (by condition of the problem)
Therefore the triangle BCF is isosceles.
By the property of an isosceles triangle:
∠CFB = ∠ABF (since these are cross-angles)
It turns out that ∠CBF = ∠ABF
Therefore, BF is a bisector.

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