Triangles within square

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Points E and F lie on the sides BC and CD of rectangle ABCD, the AEF is an equilateral triangle. point M is the midpoint of the AF. Prove that the triangle BCM is equilateral.

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Now join. See that $\angle EMF$ is $90^{\circ}$. [As $\triangle AEF$ is equilateral & $M$ is the midpoint]. $\angle ECF$ is $90^{\circ}$.

So quadrilateral $EMFC$ is cyclic. So $\angle MCB = \angle MFE = 60^{\circ}$. Similarly, quadrilateral $AMEB$ is cyclic. So, $\angle MBE = \angle MAE = 60^{\circ}$. So in triangle $MBC$, $\angle MBC = \angle MCB = 60^{\circ}$. Therefore, $MBC$ is equilateral.