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?
An analog of Mollweide’s Formula for a cyclic quadrilateral
euclidean-geometrytrigonometry
Related Solutions
Le us assume that $D$ lies at the origin and $\widehat{CDA}=\theta$. Then the coordinates of $A$ are $(x\cos\theta,x\sin\theta)$ and the line through $A$ which is orthogonal to $DA$ has equation $y(t)=-\frac{\cos\theta}{\sin\theta}t+\left(x\sin\theta+x\frac{\cos^2\theta}{\sin\theta}\right) $. It follows that the length of $CB$ is given by
$$ -\frac{\cos\theta}{\sin\theta} y + \left(x\sin\theta+x\frac{\cos^2\theta}{\sin\theta}\right)=\frac{x-y\cos\theta}{\sin\theta} $$
and by symmmetry the length of $AB$ is given by $\frac{y-x\cos\theta}{\sin\theta}$.
In equivalent terms: complete the quadrilateral, notice the relevant proportions then remove the attached pieces.
The Usual Law of Tangents
Applying the formulae for the Sum of Sines and the Sum of Cosines, we get $$ \begin{align} \frac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)} &=\frac{2\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}{2\cos\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}\\[6pt] &=\tan\left(\tfrac{A+B}2\right)\tag1 \end{align} $$ Substituting $B\mapsto-B$ in $(1)$ and then dividing by $(1)$ gives $$ \frac{\sin(A)-\sin(B)}{\sin(A)+\sin(B)}=\frac{\tan\left(\frac{A-B}2\right)}{\tan\left(\frac{A+B}2\right)}\tag2 $$ If we let $a$ and $b$ be the sides opposite angles $A$ and $B$ respectively, the Law of Sines leads to $$ \frac{a-b}{a+b}=\frac{\sin(A)-\sin(B)}{\sin(A)+\sin(B)}\tag3 $$
Equating $(2)$ and $(3)$, we get the Law of Tangents $$ \frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}2\right)}{\tan\left(\frac{A+B}2\right)}\tag4 $$ where $a$ and $b$ are the sides of a triangle opposite angles $A$ and $B$ respectively.
Inscribed Generalization
The power of the point $S$ is equal to both $\overline{SA}\,\overline{SD}=\overline{SB}\,\overline{SC}$; therefore, $$ \frac{\overline{SA}}{\overline{SB}}=\frac{\overline{SC}}{\overline{SD}}\tag5 $$ and since $\angle ASB=\angle CSD$, SAS says that $$ \triangle ASB\simeq\triangle CSD\tag6 $$ Therefore, $$ \frac{\overline{SA}}{\overline{SC}}=\frac{\overline{SB}}{\overline{SD}}=\frac ac\tag7 $$ Thus, we have $$ \begin{align} b&=\overline{SB}-\overline{SC}=\overline{SB}-\overline{SA}\frac ca\tag{8a}\\[6pt] d&=\overline{SA}-\overline{SD}=\overline{SA}-\overline{SB}\frac ca\tag{8b} \end{align} $$ From $(8)$, we can solve $$ \begin{align} \overline{SA}&=a\frac{ad+bc}{a^2-c^2}\tag{9a}\\[3pt] \overline{SB}&=a\frac{ab+cd}{a^2-c^2}\tag{9b} \end{align} $$ Now we are ready to apply the usual Law of Tangents from $(4)$: $$ \begin{align} \frac{\tan\left(\frac{\alpha-\beta}2\right)}{\tan\left(\frac{\alpha+\beta}2\right)} &=\frac{\overline{SB}-\overline{SA}}{\overline{SB}+\overline{SA}}\tag{10a}\\ &=\frac{(ab+cd)-(ad+bc)}{(ab+cd)+(ad+bc)}\tag{10b}\\[6pt] &=\frac{(a-c)(b-d)}{(a+c)(b+d)}\tag{10c} \end{align} $$
Best Answer
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
which is formula $(3)$ on MathWorld's "Newton's Formulas" entry. $\square$