Prove that it is always possible to subdivide a given trapezium into two similar trapeziums. 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. 
 A: 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 $PQ\parallel AB$.

A: I will call the length of PQ $y$ and that of AP $z$. Since the two trapeziums are similar, we must have $\frac ya=\frac by$, so $y=\sqrt{ab}$. The distance of P from A can then be calculated along the same lines:
$$\frac yz=\frac b{c-z}$$
$$y(c-z)=bz$$
$$z=\frac{yc}{b+y}=\frac{c\sqrt{ab}}{b+\sqrt{ab}}$$
To show the existence of the division asked for in the title, note that $\frac y{b+y}<1$ and therefore $z<c$; P therefore always lies on AD proper, not on the extension of it.
The line PQ can then be constructed easily, knowing that $AB\parallel PQ\parallel CD$.
