TITLE: Prove that it is always possible to subdivide a given trapezium into two similar trapeziums.
QUESTION [6 upvotes]: Given that ABCD is a trapezium with AB // DC. Let $a = AB \lt CD = b$, $DA = c$ and $BC$ be also known.

Prove that there is a line PQ (with P on DA and Q on BC) drawn parallel to AB such that (Trap ABQP) ~ (Trap PQCD).
If the title is true, then
(1) How far is P from A (in terms of a, b, c)? and 
(2) How can PQ be constructed in the Euclidean way?
Obviously, the trapeziums are equi-angular.

REPLY [3 votes]: 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.


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


As an alternative way based on the same principle,

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