Generalization of two formulae and an alternative proof of Bretschneider's formula

"If we do not succeed in solving a mathematical problem, the reason frequently consists in our failure to recognize the more general standpoint from which the problem before us appears only as a single link in a chain of related problems. After finding this standpoint, not only is this problem frequently more accessible to our investigation, but at the same time we come into possession of a method which is applicable also to related problems." — David Hilbert

 

The following formulae generalize $(1)$ in my previous post Killing three birds with one stone. For implications in a triangle see also Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle. 

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.

$\square$

An alternative form of Bretschneider's formula

We encourage readers to prove the following formula for themselves.

Prove that the area of a general convex quadrilateral is given by the following formula:

$$K=\sqrt{abcd\sin^2\left({\frac{\alpha+\gamma}{2}}\right)-s(s-c-d)(s-b-d)(s-b-c)},$$

where $a$, $b$, $c$ and $d$ are the sides lengths, $s$ is the semiperimeter, and  $\alpha$ and $\gamma$ are opposite angles.

A concept map of identities $(9)$ and $(10)$

Below you can find a concept map of the identities $(9)$ and $(10)$ so you can see clearly what's going on here (click on the image to have a better view). 

It would be interesting to investigate whether identities $(9)$ and $(10)$ can be generalized to other geometries.

I have organized my ideas presented in this page (and related links) and put it in a draft paper which you can download here.