Bretschneider's Formula

Bretschneider's formula gives the area of a quadrilateral, $\Delta$, by the following formula:

$$\Delta^{2} = (s-a)(s-b)(s-c)(s-d)-abcd\cos^{2}\left(\frac{B+D}{2}\right).$$

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Notation: Let $[X]$ represent the area of a polygon (here, a triangle or a quadrilateral). Also let $\Delta = [ABCD]$, and let $AB=a$, $BC=b$, $CD=c$, $DA=d$. Let $s$ be the semiperimeter of the quadrilateral $ABCD$, given by $s=\dfrac{a+b+c+d}{2}$.

Figure for the proof

Construction: Join $\overline{AC}$.

Proof:

We have

$$\begin{aligned} [ABCD] &= [ABC] + [ACD] \\ \Delta &= \frac{1}{2}ab\sin B + \frac{1}{2}cd\sin D \\\\ \Rightarrow \ 2ab\sin B + 2cd\sin D &= 4\Delta \qquad {\color{#3D99F6} (1)} \end{aligned}$$

Now we will use cosine rule to find $AC$ in two ways, once using $\Delta BAC$ and once using $\Delta DAC$.

$$\begin{aligned} AC^{2} &= a^{2} + b^{2} - 2ab\cos B \\ &= c^{2} + d^{2} - 2cd\cos D \\\\ \Rightarrow \ 2ab\cos B - 2cd\cos D &= a^{2} + b^{2} - c^{2} - d^{2}. \qquad {\color{#3D99F6} (2)} \end{aligned}$$

Squaring and adding equations ${\color{#3D99F6} (1)}$ and ${\color{#3D99F6} (2)}$, we have

$$\begin{aligned} 4a^{2}b^{2}+4c^{2}d^{2} - 8abcd(\cos B \cos D - \sin B \sin D) &= 16\Delta^{2} + \big(a^{2}+b^{2} - c^{2}-d^{2}\big)^{2} \\ 4a^{2}b^{2}+4c^{2}d^{2}-8abcd\cos(B+D)&=16\Delta^{2}+\big(a^{2}+b^{2}-c^{2}-d^{2}\big)^{2} \\ (2ab+2cd)^{2}-8abcd-8abcd\cos(B+D)&=16\Delta^{2}+\big(a^{2}+b^{2}-c^{2}-d^{2}\big)^{2} \\ (2ab+2cd)^{2}-8abcd\big(1+\cos(B+D)\big)&=16\Delta^{2}+\big(a^{2}+b^{2}-c^{2}-d^{2}\big)^{2}. \end{aligned}$$

Then

$$\begin{aligned} 16\Delta^{2} &= (2ab+2cd)^{2}-\big(a^{2}+b^{2}-c^{2}-d^{2}\big)^{2}-8abcd\times 2\cos^{2}\left(\frac{B+D}{2}\right) \\ &=\big(2ab+2cd+a^{2}+b^{2}-c^{2}-d^{2}\big)\times\big(2ab+2cd-a^{2}-b^{2}+c^{2}+d^{2}\big)-16abcd\cos^{2}\left(\frac{B+D}{2}\right) \\ &=\big((a+b)^{2}-(c-d)^{2}\big)\big((c+d)^{2}-(a-b)^{2}\big)-16abcd\cos^{2}\left(\frac{B+D}{2}\right) \\ &=(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)-16abcd\cos^{2}\left(\frac{B+D}{2}\right) \\ &=(2s-2d)(2s-2c)(2s-2b)(2s-2a)-16abcd\cos^{2}\left(\frac{B+D}{2}\right) \\ &=16(s-a)(s-b)(s-c)(s-d)-16abcd\cos^{2}\left(\frac{B+D}{2}\right) \\\\ \implies \Delta^{2} &=(s-a)(s-b)(s-c)(s-d)-abcd\cos^{2}\left(\frac{B+D}{2}\right), \end{aligned}$$

which is the Bretschneider's formula. $_\square$

Also see the pages on cyclic quadrilaterals and Brahmagupta's formula.