Prove that it is always possible to subdivide a given trapezium into two similar trapeziums.

It is enough to construct $PQ$ in such a way that $\frac{AB}{PQ}=\frac{PQ}{CD}$, i.e. to construct the geometric mean of $AB$ and $CD$. Here it is a possible approach, exploiting the power of a point with respect to a circle.

1. Let $E\in CD$ be a point such that $AE\parallel BC$;
2. Let $\Gamma$ the circumcircle of $ADE$ and $CF$ (with $F\in\Gamma$) a tangent to $\Gamma$;
3. Let $G\in CD$ be such that $CG=CF$;
4. Let $P\in AD$ be such that $PG\parallel BC$;
5. Let $Q\in BC$ be such that $PQ\parallel AB$.

As an alternative way based on the same principle,

1. Let $X=AD\cap BC$;
2. Let $\Gamma$ be a circle with diameter $AD$;
3. Let $T\in\Gamma$ be such that $XT$ is tangent to $\Gamma$;
4. Let $P\in AD$ be such that $XP=CT$;
5. Let $Q\in BC$ be such that $PQ\parallel AB$.