In the image, $ABCD$ is a square, $AM=ME$ and $\angle CME = 90^{\circ}$
the original problem was to prove $\alpha = \frac 32 \theta$, but I'm more concerned with the construction of the figure. If $X = AC \cap ME, H = BD \cap CM $ then $H$ is orthocenter of $\triangle CXE$ which gives us a few cool circles, but I still need to use the fact that $AM=ME$ to figure out this construction. Any ideas?
We can try to get the Co-ordinates "analytically" , to get insight on whether Construction is Possible & what Construction may work.
Short Summary : It is not Possible !
We can let $M$ move on $AD$ , then cut the arc with center $M$ having radius $AM$ to get the Point $E$ below the Square ,which indicates that $AM$ must be larger than half $AD$.
Let the Square Diagonal intersect @ the Origin to give this Image :
The Square is shifted to make Origin the Center. The Side is then scaled to make it $1+1=2$. Hence the 2 Diagonals are the lines $y=x$ & $y=-x$.
Point M will have the Co-ordinates $(M,-1)$. Point E will have the Co-ordinates $(E,-E)$. Here $M$ must be Positive & $E$ must be larger than $1$.
When $M=0$ ( at the Mid-Point ) , $\angle CME$ is less than $90$.
When $M=1$ ( at $D$ ) , $\angle CME$ is more than $90$.
Somewhere in-between , $\angle CME$ must be $90$.
Writing the Equations :
$AM=ME$ :: $(1+M)^2=(M-E)^2+(E-1)^2$
$CM \perp ME$ :: $(1+1)/(1-M) \times (1-E)/(E-M) = -1$
Via Wolfram , the Unique Solution (Domain : Positive real) is very Cumbersome :
The Core Point which we get here : The Solution has Cube roots , hence it is not Possible to Construct with Compass+Straight-Edge.