Taking the adjacent sides AB and ad of the parallelogram ABCD as one side, we make an equilateral triangle ABF and ade outside the parallelogram, and prove that △ CEF is also an equilateral triangle

It is proved that AFE, BCF and CDE are congruent triangles,
DC=AB=BF=AF,BC=AD=DE=AE
Angle ABC = angle ADC = 180 degrees - angle bad, angle ABF = angle ade = angle ead = angle BAF = 60 degrees
Therefore, angle CBF = angle CDE = angle EAF 〈 angle EAF = 360 - 60 - 60 - angle DAB = 180 - angle DAB + 60 = angle CBA + 60 = angle CBF 〉
So, de = AE = CB, DC = AF = BF, angle CDE = angle FAE = angle FBC,
So triangle CDE is equal to triangle FAE is equal to triangle FBC
So CE = EF = CF
The triangle CEF is a regular triangle

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