The following concerns the 1914 paper The Inscribed and Circumscribed Squares of a Quadrilateral and Their Significance in Kinematic Geometry of Hebbert.
Context. Hebbert presents
THEOREM I. In every quadrilateral may be inscribed at least one square having a vertex on each of the four sides. If there is more than one such square in a given quadrilateral, there is an infinite number.
The proof amounts to finding an equation by elementary trigonometry for a certain angle $\delta$ in terms of the parameters of the given quadrilateral†. He shows that $\tan\delta$ is given by some explicit ratio which does not depend on $\delta$, and thus has a real solution (which gives the first part of the theorem). For the second part of the theorem (multiple solutions imply infinitely many), he appeals to a general result of A. Hurwitz without any elaboration:
The equation [$\tan\delta={\frac\cdots\cdots}$] can have more than one solution [in $[0,\frac\pi2)$] if and only if the right-hand member is $0/0$ and then there are an infinite number of solutions, according to a theorem stated by A. Hurwitz:
"If it is possible in special cases to get more than $n$ solutions for a problem which in general has $n$ solutions determined by the roots of an equation of the $n$th degree, then in these special cases there are an infinite number of solutions."
This brings me to my confusion.
Question 1. What sense of "degree" is Hurwitz employing and why does it apply as claimed in this case to the (transcendental) tangent function (or its inverse)? Why can we guarantee infinitely many solutions in this case?
If it helps, Hurwitz's original 1879 result in German can be found here.
If it is in fact unclear what's meant in the above passage, then I must ask
Question 2. Is Hebbert's Theorem I correct? Or perhaps correct only for convex quadrilaterals? (In any case, I would welcome a more modern reference.)
† It doesn't seem to me that the details matter so much, but for completeness, the equation is $$ \tan\delta = \frac{d(\csc\beta + \cos\beta + \cot\gamma\sin\beta) - a(1+\cot\beta)}{d(\cot\gamma\cos\beta-\sin\beta)+a(1+\cot\alpha)} $$ where $a,d,\alpha,\beta,\gamma$ are fixed parameters of the given quadrilateral. Specifically, the internal angles $\alpha,\beta$ include side $d$ and $\beta,\gamma$ include side $a$. The angle $\delta$ is the internal angle along side $d$ of the triangle one of whose sides is a side of the desired inscribed square and whose other two sides are contained in the sides of the quadrilateral incident to $\alpha$.
