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.

Consider a hypothetical solution $\square ABCD$ with diagonals meeting at $X$, and with angle measures and segment lengths as shown:

Then 
$$\tan \alpha = \tan A = \tan(\alpha_1 + \alpha_2) = \frac{\tan\alpha_1+\tan\alpha_2}{1-\tan\alpha_1\tan\alpha_2} = \frac{\frac{d}{a}+\frac{b}{a}}{1-\frac{d}{a}\frac{b}{a}} = \frac{a(b+d)}{a^2-bd} \tag{1}$$
$$\tan\beta  = \frac{b(a+c)}{b^2-ac} \tag{2}$$
$$\tan\gamma = \frac{c(b+d)}{c^2-bd} \tag{3}$$
$$\tan\delta = \frac{d(a+c)}{d^2-ac} \tag{4}$$
To show that this hypothetical solution is valid, we need only solve equations (1) through (4) for $b$, $c$, $d$ in terms of $\alpha$, $\beta$, $\gamma$, $\delta$, and $a$ (which we can take to be $1$). This is do-able, and the algebra gets no more complicated than quadratics (so that the solution is constructible), but the expressions are a bit messy. I'll post more after I do some clean-up.
Edit. After considerable manipulation, (I think) the above equations reduce to these:
$$\begin{align}
  (a^2-b^2)\sin\alpha\sin\beta \cos(\gamma+\beta) + a b \left( \sin\alpha  \cos(\gamma+2\beta) + \cos\alpha\sin\gamma \right) &= 0 \\[4pt]
  (a^2-d^2)\sin\alpha\sin\delta \cos(\gamma+\delta)\;+ a d \left( \sin\alpha  \cos(\gamma+2\delta) + \cos\alpha\sin\gamma \right) &= 0 \\[4pt]
  2 (a^2+c^2) \sin\alpha\sin\gamma \cos(\gamma+\beta) \cos(\gamma+\delta)  
\qquad-   a c ( k + \sin^2(\alpha-\gamma) ) &= 0
\end{align}$$
where 
$$k := -1 + \sin^2\alpha + \sin^2\beta + \sin^2\gamma + \sin^2\delta + \cos(\alpha-\gamma) \cos(\beta-\delta)$$
Consequently, we have
$$\begin{align}
\frac{b}{a} &= -\frac{ \sin\alpha \cos(\gamma+2\beta) + \cos\alpha \sin\gamma \pm \sqrt{k}}{2\sin\alpha \sin\beta \cos(\gamma+\beta)} \\[6pt]
\frac{d}{a} &= -\frac{ \sin\alpha \cos(\gamma+2\delta) + \cos\alpha \sin\gamma \pm \sqrt{k}}{2\sin\alpha \sin\delta \cos(\gamma+\delta)} \\[6pt]
\frac{c}{a} &= \frac{ k + \sin^2(\alpha-\gamma) \pm 2 \sin(\alpha-\gamma) \sqrt{k}}{4 \sin\alpha \sin\gamma \cos(\gamma+\beta)\cos(\gamma+\delta)}= \frac{\left(\;\sin(\alpha-\gamma) \pm \sqrt{k} \;\right)^2 }{4 \sin\alpha \sin\gamma \cos(\gamma+\beta)\cos(\gamma+\delta)}
\end{align}$$
where, for now at least, resolution of the "$\pm$"s is left as an exercise to the reader.
Note that the relation $\alpha+ \beta+\gamma+\delta = 360^\circ$ causes each of these to have myriad representations. I don't claim to have given the best possible ones; in fact, I suspect there are representations that make the relations far more clear. 
As mentioned, the various quantities are constructible, since the lengths are at most as complicated as a square root. Formulation of a construction strategy will have to wait. 
Edit 2. If we normalize our lengths, say, with a constant sum,
$$a + b + c + d = s$$
then we can express each length independently. With $m := \pm\sqrt{k}$, we have
$$\frac{a}{s} = \frac{\left(\; m + \sin(\alpha-\gamma) \;\right)\left(\;m + \sin(\beta+\delta) + 2 \sin\beta\sin\delta\;\right)}{2m\left(\;2\sin(\beta+\delta)+\cos(\beta-\delta)-\cos(\alpha-\gamma) \;\right)}$$
while expressions for $b$, $c$, $d$ arise by cyclically permuting the angles, $\alpha\to\beta\to\gamma\to\delta\to\alpha$. A different normalization (for instance, $a^2+b^2+c^2+d^2=s^2$ seems a classic choice) would lead to different —potentially better— representations, but my attempts at symbol-wrangling haven't resulted in anything particularly nice. 
By the way, to verify that the earlier ratios hold, it helps to know that
$$m^2 - \sin^2(\alpha-\gamma) = 4\sin\alpha\sin\gamma\cos(\gamma+\beta)\cos(\gamma+\delta)$$
Therefore multiplying $a$ by the $c/a$ ratio above turns out to be an overly-complicated way to flip a single sign: $m + \sin(\alpha-\gamma) \;\to\; m - \sin(\alpha-\gamma)$, which matches the considerably-easier process of exchanging $\alpha$ and $\gamma$ (and exchanging $\beta$ and $\delta$, which actually does nothing) in the formula for $a$.
