Rhomboid construction by compass and straight-edge

Question

Consider the following given points:

I should construct a Rhomboid $ABCD$ using the facts that $P \in BC$, $Q \in CD$ and $\{M\} = AC \cap BD$.Thus I get

I do not really know how to proceed from here. Has anyone a hint for me?

Answer 1

Since a rhomboid is point symmetric, its diagonals bisect one another. This means point $M$ is exactly in the middle between $B$ and $D$. If you draw a circle around $M$ with radius $MB$ you get $D$ as the other point of intersection with that diameter line.

From there it should be straight-forward: Connect $D$ with $Q$ to get line $CD$, intersect with $BP$ to get $C$. Then draw some more parallels to get point $A$.