Construct quadrangle with given angles and perpendicular diagonals The following came up when I worked on the answer for a different question (though it was ultimately not used in this form):
Proposition.
Given positive angles $\alpha,\beta,\gamma,\delta$ with $\alpha+\beta+\gamma+\delta=360^\circ$, $\beta<180^\circ$, $180^\circ< \alpha+\beta<270^\circ$, $180^\circ< \beta+\gamma<270^\circ$, there exists a convex quadrangle $ABCD$ with $\angle A=\alpha$, $\angle B=\beta$, $\angle C=\gamma$, $\angle D=\delta$, having $AC\perp BD$.
Proof: Ignoring the orthognality condition, there are many possible quadrangles with the given angles that can be continuously transformed into each other. For such let $P$ denote the intersection of $AC$ and $BD$. In the degenerate case $A=B$, we get $\angle CPD=\alpha+\beta-180^\circ<90^\circ$, in the degenerate case $B=C$, we get $\angle DPA=\beta+\gamma-180^\circ<90^\circ$. 

Then the Intermediate Value Theorem guarantees the existence of a case where $\angle CPD=90^\circ$. $_\square$

My question is: Can someone provide a proof not relying on continuity arguments? That is, something more classic Greek compass-and-straightedge-y constructive?
Edit: I had to update and add $\beta<180^\circ$ to the condition in the proposition  - the old version would have allowed $\beta\ge180^\circ$ and so no convex quadrilateral at all. If we allow nonconvex quadrangles and diagonals to interesect in the exterior, this additional condition should be unnecessary.

The problem posted is equivalent to this one:
Let $a'$, $b'$, $c'$ and $d'$ rays starting from point $M$, such that $m(\angle a'Mb')=\alpha$, $m(\angle b'Mc')=\beta$, $m(\angle c'Md')=\gamma$ and $m(\angle d'Ma')=\delta$. Construct a rectangle $FGHI$ such that $F \in b'$, $G \in c'$, $H\in d'$ and $I \in a'$. See fig. 1.

Fig.1 - Rectangle that develops to the required quadrangle.
The reason: If we reflect point M across the lines $f$, $g$, $h$ and $i$ (the sides of the rectangle), we will get the quandrangle required by OP.
To solve this new problem, let's derive some equations. See fig. 2.

Fig.2 - Rectangle in a coordinate system.
Let 
$$I=(-1, \tan \alpha),$$
$$F=(-b, 0),$$
$$G=(a, a \tan \beta).$$
As line $i$ is perpendicular to line $f$, we get:
$$a+b=\frac{a \tan \alpha \tan \beta}{1-b} \quad (1),$$
or
$$a=\frac{b(1-b)}{ \tan \alpha \tan \beta -1+b} \quad (2).$$
As $x_H-x_I=x_G-x_F$, we get:
$$\tan(\alpha+\delta)- \tan \alpha=(a+b) \tan(\alpha+\delta) +a \tan \beta. \quad(3)$$
Substituting $(1)$ and $(2)$ in $(3)$, we get:
$$\tan \beta b^2 + [\tan(\alpha+\delta)- \tan \alpha-\tan\alpha \tan \beta \tan(\alpha+\delta)- \tan \beta]b+ [\tan(\alpha+\delta)- \tan \alpha](\tan\alpha \tan \beta-1)=0. \quad (4)$$
Lill's method is apropriated to solve eqution $(4)$ for $b$. The coefficients of equation $(4)$ can be easily constructed, since they are products or the results of additions/subtractions of constructible quantities.
Knowing the value of $b$ the construction of the rectangle and of the needed reflections is not difficult.
