On Fig. 21, the dashed line shows how to cut the hexagon – one has to draw a segment $CF$ or $BD$.

Let us now see where the line of separation of $ABCDEF$ can be drawn.
If it passes through a vertex of a hexagon and is different from lines $AC$ and $BE$, e.g. $AL$, then
on the side, a new point ($L$) appears, meaning the resulting polygons must have 7 vertices, two of
which are counted twice (in this case, $A$ and $L$), which means that in total, these polygons must
have 9 vertices, which is not possible for both two triangles and two quadrilaterals. Analogously,
if the line does not pass through the vertex (e.g., $MN$), then in total, there must be 10 vertices,
which is also impossible for two triangles and two quadrilaterals. The only case left is when the
segment connects two vertices of a hexagon, e.g. $BE$, then in total this yields 8 vertices, which
could be formed by two quadrilaterals. But it suffices
to check all such segments to see that none such
segment partitions the hexagon into two
quadrilaterals. $BE$ and $FD$ are the only such
segments, and each of them partitions the hexagon
into a triangle and a pentagon.