An Alternative Form of Bretschneider's Formula

If you are already familiar with Bretschneider's Formula, have you ever wonder how would it like if we interchange the cosine-part by a sine-part?

There are three different forms of expressing the Bretschneider's Formula in MathWorld. In this note we will give another one which is almost as simple as the original one.

Given a general convex quadrilateral with sides $a$, $b$, $c$ and $d$, its area is given by the formula

$$K=\sqrt{abcd\sin^2\left({\frac{\alpha+\gamma}{2}}\right)-s(s-c-d)(s-b-d)(s-b-c)}\tag{1},$$

where $s$ is the semiperimeter and  $\alpha$ and $\gamma$ are opposite angles.

The proof is based on the following unexpected simplification lemma.

Lemma 1. Given a general quadrilateral with sides $a$, $b$, $c$ and $d$, then

$$(s-a)(s-b)(s-c)(s-d)+s(s-c-d)(s-b-d)(s-b-c)=abcd,\tag{2}$$

where $s$ is the semiperimeter.

Proof. Let's focus on the left side of the identity. Substituting and rewriting as difference of squares,

$$\begin{align*}\frac{\left[(c+d)^2-(a-b)^2\right]\left[(a+b)^2-(c-d)^2\right]}{16}+\frac{\left[(a+b)^2-(c+d)^2\right]\left[(a-b)^2-(c-d)^2\right]}{16}&=\\\frac{(a+b)^2\left[(c+d)^2-(c-d)^2\right]+(a-b)^2\left[(c-d)^2-(c+d)^2\right]}{16}&=\\\frac{4cd(a+b)^2-4cd(a-b)^2}{16}&=abcd.\end{align*}$$

$\square$

Now, consider the original Bretschneider's Formula, 

$$K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\frac{\alpha+\gamma}{2}\right)}.\tag{3}$$

Using the Pythagorean Identity $\sin^2{\left(\frac{\alpha+\gamma}{2}\right)}+\cos^2{\left(\frac{\alpha+\gamma}{2}\right)}=1$ in combination with $(2)$, you get $(1)$.

Remark. Assume $d=0$. Then $(2)$ reduces to 

$$s(s-a)(s-b)(s-c)+s(s-b)(s-c)(s-b-c)=0$$

From which we get the following alternative form of Heron's Formula:

$$K=\sqrt{-s(s-b)(s-c)(s-b-c)}$$

or

$$K=\sqrt{-s(s-a)(s-c)(s-a-c)}$$

$$K=\sqrt{-s(s-a)(s-b)(s-a-b)}$$