$\textcolor{blue}{\text{ If two medians of a triangle are equal, then it is an isosceles triangle}}$
Since, $AM \cap BE= O \rightarrow \text{ centroid }, $ therefore, $OE=OM \text{ and } AO=BO $
By SAS Congruence, $\Delta ADE \cong \Delta BOM $
Hence, $AE=BM \implies AC=BC $
Using Claim 1,
Let $\angle A=\angle B=\angle CEM=\angle CME $, By angle chasing and using the fact that $CC_1 \perp EM \text{ and } \angle COE=\angle COM=x $
$$\boxed{\angle MEO=\angle EMO =90^{\circ}-x \implies \angle BEC=\angle AMC=90^{\circ}} $$Hence, the medians are also altitudes, Therefore, $\Delta ABC $ is equilateral $$AB=\frac{AM}{\sin 60^{\circ}}=\boxed{2\sqrt{3}}$$