Irska

Let $FH$, with $F$ on $AB$ and $H$ on $AC$, bisect the area of $△ABC$. Let the midpoints of $AB$ and $AC$ be $E$ and $D$ respectively and let the centroid of $△ABC$ 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 $△GDH\sim △FGB$ 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.