We first list some facts which are basic to the construction. The parallel sides $AB$ and $CD$ of the desired trapezoid $ABCD$ must both be parallel to $p$. The distance $CE$ between them is equal to the diameter of the inscribed circle. The distance $AC$ between the given points $A$ and $C$ is known. Denote the still unknown distances $AB$ and $CD$ by $a$ and $B$, respectively, and the equal distances $AD$ and $BC$ by $s$. Since the two tangents from a vertex of the quadrilateral to the inscribed circle are equal, \[ AB+CD=AD+BC \Leftrightarrow a+b=2s \] so $s=\dfrac{1}{2} \left( a+b \right)$. If $a<b$, then $AE=a+BE=\dfrac{1}{2} \left( a+b \right)=s$. If $a>b$, then $a-2BE=b$, so $BE=\dfrac{1}{2} \left( a-b \right)$ and $AE=a-BE=\dfrac{1}{2} \left( a+b \right)=s$. Thus in both cases $AE=s$. Construction: In plane $P$, constructa line $l$ through $A$ parallel to $p$; in plane $Q$, construct a line $m$ through $C$ parallel to $p$. Construct a perpendicular to $CD$ at $C$ meeting $l$ at $E$. With $AE$ as a radius and $C$ as a center, draw an arc of a circle intersecting line $AE$ at $B$. With the same radius and $A$ as a center, draw an arc of a circle intersecting line $m$ at $D$. Both desired vertices $B$ and $D$ are now located. Note: If $CE<AE$, two solutions exist. If $AE=CE$, ther is only one solution, a square. If $CE>AE$, the problem has no solution.

the above problem, and the variation in which the quadrilateral is named ABDC instead of ABCD, have been solved by Kostas Dortsios in pdf here (in English and in Greek)