Problem 6

Let $c$ be the line through $C$ parallel to $p$. For the selected $B$, project it orthogonally to $c$ and call the projection $B'$. On the directed line $c$, construct $D$ so that $\overrightarrow{A'D}=\overrightarrow{A'B'}-\overrightarrow{A'C}$; this is the standard completion of the isosceles trapezoid (the sign selects the side consistent with the vertex order). The circle through $A'$ was chosen so that $BC=A'C$, and the parallel-base completion gives $AD=BC$. Consequently $AB+DC=2A'C=AD+BC$, so the tangential-quadrilateral criterion gives an incircle. Depending on the circle intersections, there may be two solutions, one square, or none.