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.

I think ruler compass construction should be completed at first;
[Earlier, redone.The construction can be completed only upto two parameters of arbitrariness , one for scale ( naturally as only angles $ \alpha,\beta, \gamma$  and $\delta  $ are given) and another for adjoining sides ratio AB/BD at B.Draw AB and AD so that angle DAB = $ \alpha$. Construct semi circle on AB and AD as diameters intersecting at point O, where yet un-constructed diagonals would intersect at right angles.Join OB,OD, OA and extend AO on the other side towards an yet unknown point/direction C.Mark off $ \beta$ and $ \delta $ at at B and D respectively. The three lines are concurrent at C making angle $ 2\pi -( \alpha+\beta+\delta) $ A Java or Geometric SketchPad can track all solutions by changing AB/BD ratio or parameter with mouse. A static sketch would be soon uploaded. It is not necessary as an answer but for labeling. EDIT1: Regarding parametric variation, shall add a detail missing here]. 
EDIT2:
My approach is partially changed as under, only one circle is needed.

Angles at 1 and 2 are right angles in  a semi-circle. When AB is fixed and lines BC,BD are drawn, the angle sum  $ \gamma + \delta $ is a given constant. (In either position 1 or position 2 angle sum of Line L1 or L2  is the same). The inclination of variable Line L is a trig function of $ \theta $, $  S(\theta) $, which is to be worked out. For correct $\theta$ position point on semi-circle solved for between 1 and 2, Line L becomes parallel to CD.Slope of L should match that of CD.
Hope someone takes it from there, I have to go. 
