Inscribing a quadrilateral inside a rhombus Let ABCD be a rhombus, its interior angles are $\alpha<\frac{\pi}{2}$ and $(\pi-\alpha)$.
Let w, x, y, z four points located respectively in (A, B), (B, C), (C, D), (D, A).
Suppose we have as inputs the points w, z, y, z and the angle $\alpha$,
Is there a geometric method to find the points A, B, C, D such that
$\text{angle}(x,C,y)=\alpha$ ?

Thank you.
 A: No.  For instance, suppose the rhombus were actually a square with coordinates at (0,0), (0,2), (2,2), (2,0).  Suppose you are given $w,z,y,x$ as mid points of the lines of the square at (0,1), (1,2), (2,1), (1,0) and you are given that $\alpha = \pi/2$.
You might hit on the right square, but you might also fit a square by just joining up the given points.
Edit, suppose the given points were a bit closer to the true vertices.  There is more than one square that will fit.

Edit 2 (for $\alpha \ne \frac{\pi}{2}$)  You can also construct 2 congruent rhombuses (with $\alpha \ne \pi/2$) from the same points

Edit 3 (preserving the position of $\alpha$)

A: I don't know about a geometric solution, but a numerical solution can be obtained like this.
Let $m$ be some arbitrary slope and let $\theta = tan^{-1}(m)$.
Let Line$_x$ be the straight line passing through $x$ with slope $m$.   Line$_x$ is well-defined by the point it passes through and its slope.
Let Line$_y$ be the straight line passing through $y$ with slope $m' = tan(\theta + \alpha)$.   Line$_x$ and Line$_y$ intersect at point $C'$, say, and by construction the angle $(x,C', y)$ is $\alpha$.
Let Line$_z$ be the straight line passing through $z$ with slope $m$.   Line$_z$ and Line$_y$ intersect at point $D'$, say.
Let $A'$ be the point distance $C'D'$ from $D'$ (in the direction with slope $-m$).  Let $B'$ be the point distance $C'D'$ from $C'$ (in the direction with slope $-m$).  
Let $e$ be the distance of point $w$ from the line $A'B'$
Given the co-ordinates of $w$, $x$, $y$, $z$, and slope $m$, the coordinates of $A', B', C', D'$ and the distance $e$ can be calculated using linear equations and Pythagoras.
So $e$ is an explicit function of $m$, say $e = f(m)$.  Thus some numerical solver can be used to find $m*: e* = f(m*) = 0$ which will be where $w$ is on the line $A'B'$.
In some special cases(as shown in the other answer) there will be more than one solution of $f(m)=0$.
