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

 A: *

*Say $m(EAB)=\alpha$ then $m(DAF)=30-\alpha$, $m(DFA)=60+\alpha$ and $m(FEC)=30+\alpha$.

*Join $M$ and $E$.

*$m(MEA)=m(MEF)=30$, therefore $m(MEC)=60+\alpha$

*We see that $m(MEC)=m(MFD)=60+\alpha$ and $m(EMA)=m(ECF)=90$.

*As a result we can say that a circle pass through the points $M, E, C, F$.

*Finally because of the $MECF$ cyclic quadrilateral, $m(MCE)=m(MFE)=60$ 

*You can see that $MEBA$ is also a cyclic quadrilateral, therefore $m(MAE)=m(MBE)=60^\circ$ or you can joing $M$ and $D$ to see $m(MCE)=m(MBE)$

A: Set $AB=x$ and $EB=y$. You have $BC=AF\sin\left(\pi/3+\hat{EAB}\right)$, so 
$$ BC^2 = \frac{3x^2+y^2+2\sqrt{3}xy}{4}.$$
Now consider $N$ as the projection of $M$ on $AB$. We have:
$$ BM^2 = MN^2+NB^2 = AM^2 \sin^2\left(\pi/3+\hat{EAB}\right)+\left(x-AM \cos\left(\pi/3+\hat{EAB}\right)\right)^2.$$
Since $AM=\frac{AF}{2}=\frac{AE}{2}=\frac{1}{2}\sqrt{x^2+y^2}$, this leads to $BC^2=BM^2$. Since $M$ is the midpoint of $AF$, the parallel to $AB$ through $M$ intersects $BC$ in its midpoint: this gives $BM=MC$, too.
I strongly suspect that some tricky application of Ptolemy's and cosine theorem can overcome the trigonometrical part of this proof. 
