We have: $BM^2 = BD^2 + DM^2 - 2 BD . DM \cos \widehat{BDM} = \dfrac{63}{64} AB^2$ and $AM^2 = AE^2 + EM^2 = \dfrac{7}{64} AB^2$ So: $BM^2 =9 AM^2$ or $BM = 3 AM$

Hint: prove that $M$ lays on median $BP$ and $BP = 4MP$.

Let $P, Q$ be midpoints of $AC, AB$ and $F$ be midpoint of $DQ$. In parallelogram $BQPD$ : $F \in BP$ and $BF = FP$. In equilateral $\triangle ADP$ : $AE = EP$, $AM = MP$. In parallelogram $DFEP$ : $M \in FP$ and $FM = MP$. So, $AM = MP = BM/3$

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cool solution

thanks, seems complex thouh, maybe it can be simplified... EDIT: simplified. Also, another solution you can obtain by extending $AM$ on its length.

Okay, here is a solution using medians formula and cosine rule. $DE=\dfrac{CD}{2}=\dfrac{a}{4}$ - This is because we have triangle $EDC$ which has $30^\circ, 60^\circ, 90^\circ$ angles. Same thing would be with $AE=\dfrac{DA}{2}=\dfrac{b}{4}$ and $AD=\dfrac{b}{2}$ So, by medians formula $AM = \dfrac{1}{2} \sqrt{2AD^2+2AE^2-DE^2} $, so $AM = \dfrac{1}{2} \sqrt{ \dfrac{2b^2}{4}+\dfrac{2b^2}{16}-\dfrac{3b^2}{16}}$ and finally after basic transformations and using the fact that $a=b\sqrt{3}$ and $a^2=3b^2$ you will get $AM=\dfrac{b \sqrt{7}}{8}$ Using cosine rule you in triangle $BMD$ you will get this: $BM^2 = \dfrac{a^2}{4} + \dfrac{a^2}{64} - 2\dfrac{a}{2} \dfrac{a}{8} \cos 120^\circ = \dfrac{21a^2}{64} = \dfrac{63b^2}{64}$ (we used again that $a^2=3b^2$), so $BM = \dfrac{3b \sqrt{7}} {8}= 3AM$ So, that is all