A generalization of Burlet's theorem to cyclic quadrilaterals

 The Burlet's theorem is a result in Euclidean geometry, which can be formulated as follows:

Theorem 1. Consider triangle $ABC$ with $\angle{BCA}=\gamma$. Let $P$ be the point where the incircle touches side $AB$, and denote the lengths $AP$ and $BP$ as $m$ and $n$, respectively. Then

$$\Delta_0=mn\cot{\frac{\gamma}{2}},\tag{1}$$

A triangle $ABC$ with $AP=m$ and $BP=n$.

where $\Delta_0$ denotes the area of $ABC$. Note that when $\gamma=\frac{\pi}{2}$, then $\Delta_0=mn$.

We adopt unconventional notation to reduce the relation of Theorem 2 into the one referenced in $(1)$. Additionally, we denote $s$ the semiperimeter, applicable to both triangles and cyclic quadrilaterals.

Proof. Let $AB=a$, $BC=b$ and $AC=c$. By Heron's formula, we have

$$\begin{aligned}\Delta_0&=\sqrt{s(s-a)(s-b)(s-c)}\\&=\sqrt{s(s-a)(s-b)(s-c)}\color{red}{\sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}}}\\&=(s-b)(s-c)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}.\end{aligned}$$

However, using half-angle formulas for a triangle, we can express $\cot{\frac{\gamma}{2}}$ as $\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$, where $m=(s-b)$ and $n=(s-c)$. This establishes the relationship mentioned in $(1)$.

$\square$

Theorem 2 (Generalization). Consider a cyclic quadrilateral $ABCD$ with side lengths $AB=a$, $BC=b$, $CD=c$, and $DA=d$. Additionally, let $\angle DAB = \alpha$. In this context, the following relation is valid:

$$\Delta_1=(s-a)(s-d)\cot{\frac{\alpha}{2}},\tag{2}$$

A cyclic quadrilateral $ABCD$.

where $\Delta_1$ denotes the area of $ABCD$.

Proof. Similar to how we proceeded in theorem 1, using the Brahmagupta's formula, we have

$$\begin{aligned}\Delta_1&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\color{red}{\sqrt{\frac{(s-a)(s-d)}{(s-a)(s-d)}}}\\&=(s-a)(s-d)\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}.\end{aligned}$$

Now, invoking the half-angle formulas for cyclic quadrilaterals, we have that $\cot{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}$, from which the relation $(2)$ is deduced.

$\square$

Let $\angle{BCD}=\gamma$ and assume that $d=0$. We have that $\alpha=\pi-\gamma$, so substituting in $(2)$, 

When $d=0$, by the alternate segment theorem, $\alpha=\pi-\gamma$.

$$\begin{aligned}\Delta_1&=(s-a)(s-0)\cot{\left(\frac{\pi-\gamma}{2}\right)}\\&=s(s-a)\tan{\frac{\gamma}{2}}\\&=s(s-a)\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\color{red}{\sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}}}\\&=(s-b)(s-c)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\\&=(s-b)(s-c)\cot{\frac{\gamma}{2}},\end{aligned}$$

which is Burlet's theorem for a triangle since $m=(s-b)$ and $n=(s-c)$.

Remark. A. Burlet (Dublin) presented this theorem as a problem in the Nouvelles annales de mathématiques, Vol. 15, 1856, p. 290. Consequently, the theorem might have derived its name from him.