Below I present two seemingly unknown identities that I then use to provide an alternative proof of Bretschneider's formula. Made the necessary adjustments, the identities can also be used to provide alternative proofs of Brahmagupta's Formula as well as Heron's Formula (see for example https://geometriadominicana.blogspot.com/2020/07/killing-three-birds-with-one-stone.html). For implications in a triangle see also https://geometriadominicana.blogspot.com/2020/06/another-proof-for-two-well-known.html.
Here, $a$, $b$, $c$, $d$ are the sides of a general convex quadrilateral, $s$ is the semiperimeter, and $\alpha$ and $\gamma$ are two opposite angles. Then
$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)-bc\cos^2{\frac{\gamma}{2}}}{ad}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)-bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{1}$$
Proof. By the Law of Cosines,
$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{\gamma}\tag{2}$$
Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2+2bc\cos{\gamma}}{2ad}$. Now, making use of the half angle formula for cosine,
$$\begin{align*} \cos^2{\frac{\alpha}{2}}&=\frac{a^2+d^2+2ad-b^2-c^2+2bc\cos{\gamma}}{4ad}\tag{3}\\ &=\frac{a^2+d^2+2ad-b^2-c^2+2bc(1-2\sin^2{\frac{\gamma}{2}})}{4ad}\tag{4}\\&=\frac{(a+d)^2-(b-c)^2-4bc\sin^2{\frac{\gamma}{2}}}{4ad}\tag{5}\\&=\frac{(a+d+b-c)(a+d-b+c)-4bc\sin^2{\frac{\gamma}{2}}}{4ad}\tag{6}\\&=\frac{1}{ad}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)-\frac{bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{7}\\&=\frac{(s-b)(s-c)-bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{8}\end{align*}$$
$\square$
The other formula can be obtained similarly by replacing $\cos^2{\frac{\alpha}{2}}$ by $1 - \sin^2{\frac{\alpha}{2}}$ in $(3)$.
A proof of Bretschneider's formula
The formulae in $(1)$ can be rewritten as follows
$$ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d)\tag{9}$$
and
$$bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}=(s-b)(s-c)\tag{10}$$
Multiplying $(9)$ and $(10)$ we get
$$\begin{align*}\left(ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}\right)\left(bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}\right) &= (s-a)(s-b)(s-c)(s-d)\tag{11}\end{align*}$$
Expanding, factorizing, completing the squares and keeping in mind some well-known trigonometric identities,
$$\begin{align*}abcd\cos^2\left({\frac{\alpha+\gamma}{2}}\right)+\left(ad\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}+bc\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}\right)^2 &=(s-a)(s-b)(s-c)(s-d)\tag{12}\\abcd\cos^2\left({\frac{\alpha+\gamma}{2}}\right)+\left(\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\gamma}}{2}\right)^2 &=(s-a)(s-b)(s-c)(s-d)\tag{13} \end{align*}$$
Since the area of $ABCD$ can be expressed as the sum of the areas of $\triangle{ABD}$ and $\triangle{CBD}$, which in turn can be written as $\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\gamma}}{2}$, then we are done. $\square$
Are the identities (9) and (10) known? Is this proof of Bretschneider's formula known? Thanks in advance.
Edited: I have added a concept map of identities (9) and (10) so you can see clearly what's going on here.