Given any triangle $ABC$, we have to find the point $P$ inside the triangle, such that any line through $P$ divides the triangle into two pieces which are as equal as possible. We take $P$ to be the centroid $G$. We show first that in that case the smallest piece is at least $\frac{4}{9}$ of the total area. If we take a line through $G$ parallel to one of the sides, then the triangular piece has area $\frac{4}{9}$ of the total area.

Take any other line through $G$, say $RS$ as shown above. We claim that $RG > GS$. Take $S'$ to be the point obtained by rotating $S$ through $180^\circ$ about $G$. Then $GSE$ and $GS'D$ are congruent and $SE$ is parallel to $S'D$. So $R$ cannot coincide with $S'$ (because the lines $SE$ and $RD$ are not parallel, they meet at $A$). If $GR < GS'$, then $RD$ and $SE$ will meet on the wrong side of $DE$. So we must have $GR > GS'$ and hence $GR > GS$. The triangles $GRD$ and $GSE$ have the same height ($GD \sin \angle DGR$), so area $GRD >$ area $GSE$. Hence area $ARS >$ area $ADE = \frac{4}{9}$ area $ABC$.

Let $N$ be the midpoint of $AC$. As we move $R$ towards $B$, $S$ moves towards $A$. When $R$ reaches $B$, $S$ reaches $N$. So $S$ lies between $E$ and $N$. Now consider the triangles $BGR$, $NGS$. $BGR$ has the larger base, because $BG = 2GN$, and the larger height because $RG > SG$, so $RG \sin \angle RGB > SG \sin \angle SGN$. So area $BGR >$ area $NGS$ and hence area $ABN >$ area $ARS$. So area $ARS < \frac{1}{2}$ area $ABC$. Thus $ARS$ is the smaller piece but its area is bigger than $\frac{4}{9}$ area $ABC$. It remains to show that no other choice of $P$ is better than $G$.

Take lines through $G$ parallel to the sides. That gives three overlapping triangles which cover $ABC$. If $P$ is not at $G$, then it must lie inside at least one of these triangles, say $ADE$. Now take a line $D'E'$ through $P$ parallel to $DE$. Then area $AD'E' <$ area $ADE = \frac{4}{9}$ area $ABC$, so $P$ is a worse choice than $G$ (from $A$'s point of view).