Let $FH$, with $F$ on $AB$ and $H$ on $AC$, bisect the area of $\triangle ABC$. Let the midpoints of $AB$ and $AC$ be $E$ and $D$ respectively and let the centroid of $\triangle ABC$ be $G$. Assume $FH$ passes through $G$ and is not a median, i.e. $F \ne E$ and $H \ne D$.

Since $CE$ bisects the area of $\angle ABC$ then the area of $BFHC$ is equal to the area of $BEC$, hence the area of $GHC$ is equal to the area of $FEG$. This means that
$$\frac{1}{2}|GH| \cdot |GC| \sin \angle HGC = \frac{1}{2}|FG| \cdot |EG| \sin \angle FGE,$$
hence
$$|GH| \cdot |GC| = |FG| \cdot |EG| = \frac{1}{2} |FG| \cdot |GC|,$$
the last equality because of $|EG| = \frac{1}{2}|GC|$. This implies $|GH| = \frac{1}{2}|FG|$. Because $|GD| = \frac{1}{2}|BG|$ we get $\triangle GDH \sim \triangle FGB$ and so $\angle GDH = \angle FBG$ hence $AB \parallel AC$. This is impossible so $FH$ cannot bisect the area of $\angle ABC$ unless it is a median.