smc

Let $[NME]=a$ and $[BMD]=b$. The altitudes and bases of $△ANM$ and $△NME$ are the same, so $[NME]=[ANM]=a$. Since the base of $△BEM$ is twice as long as $△NME$ and the altitudes of both triangles are the same, $[BEM]=2a$. Since the altitudes and bases of $△BMD$ and $△CMD$ are the same, $[BMD]=[CMD]=b$. Additionally, since $△AME$ and $△EMB$ have the same area, $△AMC$ and $△BMC$ also have the same area, so $[AMC]=2b$. Also, the altitudes and bases of $△ABD$ and $△ADC$ are the same, so the area of the two triangles are the same. Thus, $$a+a+2a+b=2b+b$$ $$4a=2b$$ $$2a=b$$ Thus, the area of $△ABC$ is $12a$, so the area of $△NME$ is $\frac{1}{12}$ the area of $△ABC$. The answer is $\boxed{\frac{1}{12}}$.