In a possible attempt to explain a), let us focus solely on a single angle, say angle $A$. Similarly, draw tangent lines extending from the two adjacent sides, namely $AB$ and $AD$. Assuming $A\neq180$ (which we can, because it would cause $ABCD$ to be a triangle), $AB$ and $AD$ are not parallel.
This means that they meet at $A$ and continue, getting further apart as they go. If $A \lt 180$, meaning $ABCD$ is convex, $AB$ and $AD$ continue away from the shape, not intersecting any sides.
However, if $A \gt 180$, $AB$ and $AD$ enter the interior or $ABCD$ after intersecting at $A$. As the lines are infinite and the quadrilateral is not, the lines must at some point leave the shape. As two lines can only meet at a single point, and will not intersect themselves, they must leave the shape through one of the other two sides (Note Pasch's Theorem).
As both $AB$ and $AD$ are equally dependent on the angle of $A$, it is not possible for only one of the two lines to split one of the other sides.
This is based on the Inscribed Angle Theorem, which says that an inscribed angle measures half the angle of the arc that it subtends on the circle. The following proof is fairly simple.
Let $D$ be the center of the circle. Note that $\angle DAC=\angle DCA$ and that
$$
\angle DAC+\angle DCA+\angle ADC=\pi=\angle ADC+\angle CDE\tag{1}
$$
$\hspace{3.2cm}$
Subtracting $\angle ADC$ from both sides of $(1)$ and remembering that $\angle DAC=\angle DCA$, we have
$$
\angle CDE=2\angle DAC
$$
Similarly,
$$
\angle BDE=2 \angle DAB
$$
Taking a sum or difference, we get that
$$
\angle CDB=2\angle CAB
$$
$\square$
In your diagram, this means that $\angle ACB$ is half the arc from $A$ to $B$. The same is true of $\angle ADB$.
To subtend the same arc of a circle, the vertices of the angles must be on the same side of the chord between the ends of the arc. To prove the converse, we need to assume that the points considered be on the same side of the chord.
Suppose we have a point $C$ on a circle with a given $\angle ACB$. Suppose that $\angle ADB$ is the same angle, but $D$ is not on the circle.
$\hspace{3.2cm}$
Find the point $E$ at the intersection of $\overline{BD}$ and the circle. We know by the Inscribed Angle Theorem, that $\angle AEB=\angle ACB$. However, if $D$ is outside the circle, $\angle AEB\gt\angle ADB$, and if $D$ is inside the circle, $\angle AEB\lt\angle ADB$. Therefore, $D$ must be on the circle.
$\square$
Thus, if two points are on the same side of the chord, and the chord subtends the same angle at both, then they are on the same circle with that chord.
Best Answer
The source appears to be Bretschneider, C. A. Untersuchung der trigonometrischen Relationen des geradlinigen Viereckes. Archiv der Math. 2, 225-261, 1842.
The equation you're asking about is the first equation of group 32 on pg 237.
The terms are defined on pg 227, except for $\psi$, which is defined on pg 234.
The paper presents a large number of formulae in a similar vein that relate to the so-called complete quadrangle.