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.  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.

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.

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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 $PQ\parallel AB$.



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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 $PQ\parallel AB$.