Lateral Face of Prismatoid is Triangle, Trapezium or Parallelogram

Theorem

Let $\PP$ be a prismatoid.

Let $\FF$ be a lateral face of $\PP$.

Then $\FF$ is one of the following:

Suppose that $\FF$ were a polygon with more than $4$ sides.

Then at least $2$ of those adjacent sides, $S$ and $T$ say, are also sides of $\FF$.

But as $S$ and $T$ are not in the same plane as the lateral edges that form the other sides of $\FF$, they cannot form a polygon.

So the lateral faces can have no more than $4$ sides.

So, let $A$ and $B$ be the bases of $\PP$.

They are parallel by definition.

Let $L_1$ and $L_2$ be sides of $A$ and $B$ respectively which form the sides of one of the lateral faces $\FF$ of $\PP$.

Let $L_1$ and $L_2$ be such that they are not of the same length.

Then $\FF$ is a quadrilateral which has $2$ unequal sides parallel.

Hence by definition it is a trapezium.

$\blacksquare$