Solution:

Since the areas of $\triangle BON$ and $\triangle COM$ are equal we see that the areas of triangles $\triangle BCN$ and $\triangle CBM$ are also equal. Since these two triangles share the side, they must have the corresponding altitudes equal. Hence the length of perpendiculars from $M$ and $N$ to $BC$ are equal, implying that $NM \parallel BC$. Thus $MN$ is the midsegment of $\triangle ABC$ and consequently $N$ is the midpoint of $AB$.