This problem appeared in my mind when I was working on another problem, I found it interesting but still don't know how to solve it yet. So i decided to post it here, I hope we can discuss and you guy will help me to solve it.
Let $ABCD$ be a quadrilateral with $AB$ is not parallel to $CD$. Using compass and ruler, construct a point $M$ such that two triangle $AMD$ and $BMC$ are similar.
My idea is drawing the angle bisectors of $\widehat{AMB}, \widehat{BMC}, \widehat{CMD}, \widehat{AMD}$ which cut $AB, BC, CD, DA$ at $X, Y, Z, T$ respectively, then $XZ\perp YT$ at $M$. We also have, $\dfrac{XA}{XB}=\dfrac{ZD}{ZC}; \dfrac{YB}{YC}=\dfrac{TA}{TD}$.
Still have no idea what to do next :D
So, please help me. Thanks.
The ratio $\frac{AD}{BC}$ fixes the scale between the two triangles. So you are looking at a point $M$ such that
$$\frac{AM}{BM} = \frac{DM}{CM} = \frac{AD}{BC}$$
The last of these fractions is given from your quadrilateral. The other two restrict $M$ to lie on a certain Apollonian circle. So I'd construct these two circles, and find $M$ as their point of intersection.
Note that the triangles are not really similar since they have opposite orientation. Your question seems a bit vague in terms of orientation, hence I took the liberty of interpreting it like this. If you have more strict and different orientation requirements, please make them explicit.