Take a quadrilateral $ABCD$ and consider a line parallel to each of its side such that: 1- the distance between the parallel line and the side is some fixed amount $x$ and 2- the parallel line are "outside" of the quadrilateral (this is a clumsy formulation when the quadrilateral is concave, but I think the idea is clear).
These lines define a new quadrilateral $A'B'C'D'$ where $A'$ is the intersection of the parallels to $AB$ and $AD$, and likewise for the others.
Question: Assume that for all $x$, $A'B'C'D'$ is similar $ABCD$. Is $ABCD$ a tangential quadrilateral?
Note that if $ABCD$ is a tangential quadrilateral, then $A'B'C'D'$ is similar to $ABCD$ for all $x$. Indeed, up to similarity, one can always assume a tangential quadrilateral has an incircle of radius one, and this circle touches the quadrilateral at a point which is at 0°. So, up to similarity, what defines a tangential quadrilateral is the location (say in degrees) of the other 3 points on the circle where the tangents touch. Upon preforming the transformation, the point of contact remain at the same angles (on a larger circle).
My question asks for the converse of this observation.