Killing three birds with one stone

In this note we derive Heron's formula, Brahmagupta's formula and the bicentric quadrilateral's area formula, $\sqrt{abcd}$, from two formulae involving sine, cosine, semiperimeter and the side lenghts of a cyclic quadrilateral. 

Let $ABCD$ be a cyclic quadrilateral with $AB=a$, $BC=b$, $CD=c$, $DA=d$ and $s=\frac{a+b+c+d}{2}$. If $\angle{BAD}=\alpha$, then

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{ad+bc}\tag{1}$$

or 

$$sin^2{\frac{\alpha}{2}}=\frac{(a+b+c-d)(-a+b+c+d)}{4(ad+bc)}\quad and \quad cos^2{\frac{\alpha}{2}}=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\tag{2}$$

Proof. First we will find an expression for $\cos{\alpha}$ in terms of $a$, $b$, $c$ and $d$. Let $\angle{BCD}=\gamma$. By the Law of Cosines and keeping in mind that $\alpha$ and $\gamma$ are supplementary, we have

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{(180^\circ-\alpha)}\tag{3}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}$. Now, making use of the half angle formula for cosine,

$$\begin{align*} \cos^2{\frac{\alpha}{2}}&=\frac{2ad+2bc+a^2+d^2-b^2-c^2}{4(ad+bc)}\tag{4}\\ &=\frac{(a+d)^2-(b-c)^2}{4(ad+bc)}\tag{5}\\&=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\tag{6}\\&=\frac{1}{ad+bc}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)\tag{7}\\&=\frac{(s-b)(s-c)}{ad+bc}\end{align*}$$

$\square$

The other formulae can be obtained similarly by replacing $\cos^2{\frac{\alpha}{2}}$ by $1 - \sin^2{\frac{\alpha}{2}}$.

A generalization of $(1)$ together with a proof of Bretschneider's formula can be found here.

Remark. $(1)$ appears as exercise 400 in V. Panagiotis' 1000 General Trigonometry Exercises, Volume B.

A proof of Heron's Formula

For a triangle, if in $(1)$ we assume $c=0$, then we have

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{s(s-b)}{ad}\tag{8}$$

Let $\Delta_0$ be the area of $\triangle{ABD}$. Making use of the double-angle identity for sine we have

$$\sin{\alpha}=2\sqrt{\frac{s(s-b)}{ad}}\sqrt{\frac{(s-a)(s-d)}{ad}}=2\frac{\sqrt{s(s-a)(s-b)(s-d)}}{ad}\tag{9}$$

Since $\Delta_0=\frac{ad\sin{\alpha}}{2}$, it follows 

$$\Delta_0=\sqrt{s(s-a)(s-b)(s-d)}\tag{10}$$

$\square$

For more implications in a triangle see here.

A proof of the Brahmagupta's formula

Denote $\Delta_1$ the area of the cyclic quadrilateral, $ABCD$. Then

$$\begin{align*}\Delta_1&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\tag{11}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\tag{12}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\tag{13}\\&=\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-b)(s-c)}{ad+bc}}(ad+bc)\tag{14}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\tag{15}\end{align*}$$

$\square$

A proof of the bicentric quadrilateral's area formula

Since $a+c=b+d$ in a bicentric quadrilateral, the formulae in $(2)$ reduce to 

$$sin^2{\frac{\alpha}{2}}=\frac{bc}{ad+bc}\quad and \quad cos^2{\frac{\alpha}{2}}=\frac{ad}{ad+bc}\tag{16}$$

Assume $ABCD$ is a bicentric quadrilateral and let $\Delta_2$ be its area, then

$$\begin{align*}\Delta_2&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\tag{17}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\tag{18}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\tag{19}\\&=\sqrt{\frac{bc}{ad+bc}}\sqrt{\frac{ad}{ad+bc}}(ad+bc)\tag{20}\\&=\sqrt{abcd}\tag{21}\end{align*}$$

$\square$