An analog of Mollweide's Formula for a cyclic quadrilateral Note that as with Mollweide's Formulas, this version for a cyclic quadrilateral not only relates the four sides to the four angles, but also includes the angle between the diagonals. It is not clear to me if this expression generalizes the Mollweide's Formulas or Newton's Formulas (which is my intention), so my question is: does the Mollweides's Formula (or Newton's version) follows from the relation in the image?
 A: Anticipating reducing the formula to one of Newton's, rename elements of the figure as shown:

That is, angles $\alpha$ and $\gamma$ have been sub-divided into $\alpha+\alpha'$ and $\gamma+\gamma'$, and the segment labels have been scrambled: $a:=|BC|$, $b := |AD|$, $c:=|AB|$, $d:=|CD|$. Note that $\delta = \pi-\beta$ (as $\beta$ and $\delta$ are inscribed angles subtending opposing arcs). With these changes, the formula in question becomes
$$\frac{c+d}{a+b}\;\cot\frac12\theta = \frac{\sin\frac12(\alpha+\alpha'+\beta)}{\cos\frac12(\gamma+\gamma'-(\pi-\beta))} \to
\frac{c+d}{a+b}\;\cot\frac12\theta = \frac{\sin\frac12(\alpha+\alpha'+\beta)}{\sin\frac12(\gamma+\gamma'+\beta)} \tag{1}$$
Now, if we slide vertex $D$ along the circle to coincide with $C$ (so that $E$ does as well), we find that $\alpha'$ and $d$ shrink to nothing; $\gamma'$ and $\theta$ adjust to match $\beta$ and $\gamma$, respectively; and $\beta$ and $\delta$ don't change at all.

So, $(1)$ becomes
$$\frac{c+0}{a+b}\;\cot\frac12\gamma = \frac{\sin\frac12(\alpha+0+\beta)}{\sin\frac12(\gamma+\beta+\beta)} \quad\to\quad
\frac{c}{a+b}\;\cot\frac12\gamma = \frac{\sin\frac12(\alpha+\beta)}{\sin\frac12(\gamma+2\beta)} \tag2$$
Since $\alpha+\beta+\gamma=\pi$, we have
$$\alpha+\beta=\pi-\gamma \qquad\qquad\gamma+2\beta = \pi-(\alpha-\beta)$$
whence $(2)$ becomes
$$\frac{c}{a+b}\;\frac{\cos\frac12\gamma}{\sin\frac12\gamma} = \frac{\sin\left(\frac\pi2-\frac12\gamma\right)}{\sin\left(\frac\pi2-\frac12(\alpha-\beta)\right)} = \frac{\cos\frac12\gamma}{\cos\frac12(\alpha-\beta)} \tag3$$
and we can write

$$\frac{a+b}{c} = \frac{\cos\frac12(\alpha-\beta)}{\sin\frac12\gamma} \tag{$\star$}$$

which is formula $(3)$ on MathWorld's "Newton's Formulas" entry. $\square$
