TITLE: Construct quadrangle with given angles and perpendicular diagonals
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: 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.