Solution:
Let $L$ be the midpoint of $BC$. Let $BC = 2a$, $AC = 2b$, and $AB = 2c$. Applying the parallelogram law to $CLMN$ gives
$$CM^{2} + c^{2} = a^{2} + b^{2} + a^{2} + b^{2}$$
or
$$CM^{2} = 2a^{2} + 2b^{2} - c^{2}.$$
Squaring both sides of the given equation and substituting yields
$$\frac{2a^{2} + 2b^{2} - c^{2}}{4b^{2}} = \frac{3}{4}$$
which simplifies to
$$b^{2} + c^{2} = 2a^{2}$$
This equation is equivalent to $CM / AC = \sqrt{3} / 2$. Because it is symmetric in $b$ and $c$, we conclude that it is also equivalent to $BN / AB = \sqrt{3} / 2$.