I fully understand the reasons why we believe that given the four interior angles of a quadrilateral and one of its sides, it is not possible to determine the other sides (see discussion here). However, I have been playing around with the generalized Lami's theorem and have gotten some results that seem to contradict this belief. That is, it seems that some quadrilaterals can be determined given the four interior angles and one side.
Suppose we have a quad and we know all the interior angles and a side. Since we can treat distances as vectors and a quad is a close polygon, we can write $\Sigma{D_x}=0$, $\Sigma{D_y}=0$, where $D_x$ and $D_y$ stand for the horizontal and vertical components of the vectors, respectively. Another equation is given by the generalized Lami's theorem
$$D_1D_4\sin{\alpha}+D_2D_3\sin{\gamma}=D_1D_2\sin{\beta}+D_3D_4\sin{\delta}\tag{1}$$
Here is the situation that is driving me crazy. Having said all of the above, we will have the following system of equations accompanied by its graphic representation. $$\begin{equation} \Sigma{D_x}=0\quad (plane) \\\Sigma{D_y}=0\quad (plane)\\D_1D_4\sin{\alpha}+D_2D_3\sin{\gamma}=D_1D_2\sin{\beta}+D_3D_4\sin{\delta}\quad (hyperbolic\quad paraboloid) \end{equation}$$ Suppose these three graphics share some common points $(x_i, y_i, z_i)$, which are the solutions of our system of equations. Also suppose there is a solution such that $x, y, z$ are real positive numbers, does this mean that we have a determined quadrilateral? I have followed this approach to solve a quadrilateral given its four interior angles and one side. The equations give a positive solution that coincides with the scaled graph that I have built in Geogebra. Do we have a paradox here?
This is the problem that has inspired my question.
Edition to respond to Oscar. I think I've found a “counterexample” to Oscar's argument. Your argument with the parallels does not work if, in your graphic, quadrilateral ABCD is tangential. In other words, if in your graph ABCD is tangential and C', D' are points on lines AC and AD respectively, so that C'D' is parallel to CD, then ABC'D' is not necessarily tangential. At this point you might say, "well, ok, but the fact that it's tangential adds more information than just 4 angles and a side." However, if there is a relation among the 4 interior angles that characterizes the tangential quadrilaterals, then I could assign a studend the problem of determining the 3 unknown sides of a quadrilateral given 4 angles and one side, and it will depend on his knowledge to determine if this quadrilateral is tangential.
Suppose $d_1$, $d_2$, $d_3$ are the three unknown sides of a quadrilateral and $a$ is the known side. Since distances can be treated as vectors and we have a closed quadrilateral, we can generate two equations with three unknowns, namely $\Sigma{D_x}=0$, $\Sigma{D_y}=0$, where $D_x$ and $D_y$ are the horizontal and vertical components of the distances, respectively. In order for this quadrilateral to be determined, we need another equation that relates the three unknown distances, but this equation is given by Pitot's theorem in case the quad is tangential. According to the Pitot’s theorem, the two pairs of opposite sides in a tangential quadrilateral add up to the same total length, which equals the semiperimeter of the quadrilateral. Thus, our third equation would be $d_1+d_3=d_2+a$, which together with the other two equations, should generate a system with a unique solution.
Iosifescu gave a characterization for tangential quadrilaterals that only relates angles, but these angles are related to a diagonal. If a relation between the 4 interior angles could be achieved that would characterize a tangential quadrilateral, then I would be entitled to assign the problem of determining the unknown sides of a quadrilateral given 4 angles and one side.
I think the reason why my equations have given solutions that match my scaled computer graph is because I have unconsciously constructed a quadrilateral whose characteristics are essentially those of a tangential quadrilateral.
Conclusion. Is it possible to solve a quadrilateral given 4 angles and one side? If there is a relationship among the 4 interior angles that characterizes tangential quadrilaterals, then there are some quadrilaterals that can be solved given only 4 angles and one side. It will be up to you to know what kind of quad we are talking about.
The problem (in Euclidean geometry) is that the anglescare not independent. If three interior angles are given, then the fourth one must be $360°$ minus the sum of the other three. So "four" angles and a side are really only four independent parameters and you need five to specify a quadrilateral geometrically.
What happens if you don't obey this condition? Dan describes what happens with rectangles, but any quadrilateral shape presents a similar problem. In the drawing below, A and B have been fixed and so are the angles at the ends of that segment, so C and D lie on the other two sides of the blue triangle as shown. But the side connecting C with D may be any of a series of parallel lines, colored brown, giving many quadrilaterals with all angles as well as one side identical.
Off on a tangent
If we add the constraints that the quadrilateral is to be tangential -- all four sides tangent to a common circle inscribed in the quadrilateral -- then that becomes the additional constraint that enables unique solution of the quadrilateral. In terms of the drawing above, the unique circle tangent to the three sides of the blue triangle is tangent to only one of the set of parallel brown lines passing through the triangle, so only one of the quadrilaterals is tangential.
For tangential quadrilateral $ABCD$ define $AB=a,BC=b, CD=c,DA=d$. We assume that all angles are fixed and add up to 360°, and the $a$ is given among the side lengths.
If we draw diagonal $AC$ and apply the Law of Cosines to the two triangles thus formed, we find that for both triangles to share the common diagonal with the same length we need
$a^2+b^2-2ab\cos\angle B = c^2+d^2-2cd\cos\angle D.$
We also require for a tangential quadrilateral:
$a-b=d-c.$
If we square both sides of the second equation and subtract the Law of Cosines relation, all the squared terms cancel out and we are reduced to a formula that can be neatly rendered in terms of half-angles:
$ab(1-\cos\angle B)=cd(1-\cos\angle D)$
$ab\sin^2(\frac12\angle B)=cd\sin^2(\frac12\angle D)*$
We may render the analogous relation for the other two angles:
$ad\sin^2(\frac12\angle A)=bc\sin^2(\frac12\angle C)*$
If we multiply the starred equations we eliminate $b$ and $d$ and thus can solve for $c$ ($a$ is already given). This gives
$c=\dfrac{a\sin(\frac12\angle A)\sin(\frac12\angle B)}{\sin(\frac12\angle C)\sin(\frac12\angle D)}.$
With that side known, either of the starred equations gives the ratio of $b$ to $d$, from which the separate quantities are obtained using the tangential-quadrilateral relation $b+d=a+c$.