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

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

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

La importancia de tener un amigo matemático

La mayoría de nosotros, enfrentados ante un problema matemático de esos de “cuánto tardarían dos trenes en chocar si uno sale de la estación a las tal y tal…”, pensamos en nuestra juventud que de qué demonios nos serviría eso en la vida, al menos si no queríamos ser ingenieros. Hasta a un servidor, que siempre le gustó la ciencia de los números, le parecía una pérdida de tiempo averiguar la hora del trágico desenlace sabiendo que alguna señal impediría que los convoyes ocuparan la misma vía al mismo tiempo. Para eso ya habría expertos, mientras que los de letras nos dedicábamos a otras cosas. Pero conforme crecemos, nos damos cuenta de que muchas de las situaciones mencionadas en los exámenes sí se presentan en nuestras vidas, y aunque no lo hagan, las matemáticas son necesarias para prácticamente cualquier carrera que decidamos seguir, aunque sea para calcular los impuestos que debamos pagar al insaciable estado o para que no nos hagan trampa en algún negocio. Algunos más listos, como los señores que inventaron Google, supieron sacarle jugo a los números, sus ecuaciones y algoritmos para forrarse superlativamente; otros aprendieron a usarlos para ganar la lotería, apostando a lo seguro.

Probablemente los nombres de Charles Marie de la Condamine y Francois-Marie Arouet no digan nada a la mayoría, pero las cosas cambian si os digo que este segundo utilizó en su vida decenas de seudónimos, entre ellos el de Voltaire, uno de los pensadores más célebres de la Ilustración y un crítico constante de las autoridades francesas, su carencias y abusos en las décadas previas a la Revolución. Por su parte, de la Condamine no alcanzó la fama de Voltaire en los libros de historia, pero sí en los de matemáticas y en los de geografía, a través de sus valiosas investigaciones en Sudamérica, patrocinadas, como no, por sus ganancias con la lotería. Esta había sido instituida en la segunda década del siglo XVIII con la intención de revalorizar los bonos del tesoro expedidos anteriormente y, de paso, conseguir fondos nuevos para el estado. El problema es que el encargado de diseñarla, Le Pelletier Desforts, Vice Ministro de finanzas, cometió un grave error.

Los futuros socios de la Condamine y Voltaire se conocieron en una cena cuando el segundo acababa de volver a París después de uno de sus múltiples exilios. Sus finanzas no eran muy boyantes y el matemático aprovechó para contarle el fallo que había encontrado en la lotería. Él no tenía problemas de dinero, pero quería probar su teoría y sabía que Voltaire contaba con ciertas conexiones que les ayudarían en la empresa. El truco era el siguiente: todos los dueños de bonos podían comprar un billete de la lotería, uno por bono, sin importar su valor, al precio de una milésima del valor del bono. El premio equivaldría al valor del bono, más 500.000 libras extras. Así, los que comparaban billetes a 1 libra con bonos de 1000 libras, tenían la misma oportunidad de ganar el premio que alguien que había comprado el billete a 100 libras por tener un bono de 100.000 libras. Lo único que tenían que hacer los socios era comprar la mayoría de los bonos más baratos, de los que había más, y así se aseguraban ganar un premio invirtiendo menos. No todo era fácil, pues se necesitaba un capital para empezar, y aquí fue donde las conexiones sociales de Voltaire entraban en juego, sino que los billetes podían comprarse exclusivamente a través de un notario. El filósofo no tuvo problemas en encontrar uno que aceptara ser parte del cartel.

Así, cada mes Voltaire acudía a su amigo el notario a comprar los billetes y poco después volvía a cobrar sus premios. El truco funcionó durante varios meses en los que todos los participantes se llevaron pingües beneficios a costa del estado, hasta que fueron descubiertos. Resulta que la gente de aquella época acostumbraba escribir mensajes en cada cupón de la lotería, normalmente deseándose suerte a sí mismos, pero Voltaire no pudo evitar burlarse del mismo gobierno que le estaba haciendo rico. En muchas ocasiones, escribió en sus billetes cosas como “¡A la buena idea de Marie de la Condamine!”, o incluso burlas directas, “¡Brindo por el Vice-Ministro de Finanzas!”. Este, que ya se había dado cuenta de que un mismo grupo de personas ganaba casi todos los sorteos, no tardó en descubrir su identidad. Desforts demandó a Voltaire y a de la Condamine por fraude, pero como en realidad no habían hecho nada ilegal, el tribunal los exoneró y pudieron quedarse lo ganado. Eso sí, el ministro tuvo la sensata idea de cancelar la lotería.

Charles Marie de la Condamine, miembro de la Academia de las Ciencias, utilizó sus ganancias para pagarse un par de expediciones a Sudamérica, donde confirmó la teoría de Isaac Newton de que la Tierra no era completamente esférica, sino ligeramente achatada y abultada a la altura del Ecuador. También fue el primero en conocer el caucho, y quien lo introdujo a Europa junto con un documento científico demostrando sus propiedades, y realizó el primer mapa topográfico basado en observaciones astronómicas de la Cuenca del Amazonas. Por su parte, Voltaire, rico de por vida con el medio millón de libras obtenido de la lotería, se dedicó a lo que mejor sabía hacer, criticar al gobierno y promover las libertades de expresión y religión y la separación de estado e iglesia en más de dos mil libros y folletos y 20.000 cartas que aún se preservan. De la Condamine y Voltaire probablemente no utilizaron los medios más éticos para enriquecerse, pero al menos dieron buen uso al dinero.


Fuente: Ciencia Histórica

The product AI*BI*CI

Consider a triangle $\triangle{ABC}$ and its Incenter, $I$. Denote $R$ and $r$ the circumradius and inradius, respectively. Also let $AI=k$; $BI=l$; $CI=m$. Then, the following identity holds

$$klm=4Rr^2$$

Proof. We make use of the semiperimeter-half-angle formula, 

$$\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}\tag{1}$$ 

where $s$ is the semiperimeter and $\gamma$ denotes the angle $\angle{ACB}$. The proof for this formula can be found here.

Notice that $\cos{\frac{\gamma}{2}}=\frac{(s-c)}{m}$. Also, because of $(1)$ we have $\cos{\frac{\gamma}{2}}=\sqrt{\frac{s(s-a)}{ab}}$. Equating both expressions and solving for $m^2$, 

$$m^2=\frac{ab(s-c)}{s}$$

Similarly you get $k^2=\frac{bc(s-a)}{s}$ and $l^2=\frac{ac(s-b)}{s}$. Hence, 

$$(klm)^2=\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{s^3}=\frac{a^2b^2c^2s(s-a)(s-b)(s-c)}{s^4}$$

Substituting from Heron's formula,

$$(klm)^2=\frac{a^2b^2c^2\Delta^2}{s^4}$$

Simplifying and using the well-known formulas $abc=4R\Delta$ and $\Delta=rs$ you get the desired result. 

$$klm=\frac{abc\Delta}{s^2}=\frac{4R\Delta^2}{s^2}=4Rr^2$$

Still very Bretschneider, isn't it?

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

$$\begin{align*} K&=\frac{1}{4}\sqrt{4p^2q^2-(b^2+d^2-a^2-b^2)^2}\tag{1}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)-\frac{1}{4}(ac+bd+pq)(ac+bd-pq)}\tag{2}\end{align*}$$

where $p$ and $q$ are the diagonal lengths and $s$ is the semiperimeter.

In MathWorld the American mathematician Julian Coolidge is credited with giving the second form of this formula, stating "here is one [formula] which, so far as I can find out, is new," while at the same time crediting Bretschneider and Strehlke with "rather clumsy" proofs of the related formula

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

where $\alpha$ and $\gamma$ are two opposite angles of the quadrilateral.

The author of this note has realized that Coolidge's Formula is a easy consequence of Bretschneider's results (i.e., the Bretschneider's formula and Bretschneider's generalization of Ptolemy's Theorem).

In 1842 Bretschneider derived the following generalization of Ptolemy's theorem, regarding the product of the diagonals in a convex quadrilateral.

Theorem 1 (Bretschneider). Given a general  convex quadrilateral with sides of lengths $a$, $b$, $c$, and $d$, then

$$p^2q^2=a^2c^2+b^2d^2-2abcd\cos{(\alpha+\gamma})\tag{4}$$

where $p$ and $q$ are the diagonal lengths and $\alpha$ and $\gamma$ are two opposite angles of the quadrilateral.

We avoid proving Theorem 1; however, you can consult $[1]$ for that purpose. 

Using the cosine double angle formula and substituting in $(4)$,

$$\begin{align*}p^2q^2&=a^2c^2+b^2d^2-2abcd\left[2\cos^2{\left(\frac{\alpha+\gamma}{2}\right)}-1\right]\tag{5}\\&=(ac+bd)^2-4abcd\cos^2{\left(\frac{\alpha+\gamma}{2}\right)}\tag{6}\end{align*}$$

Substituting from Bretschneider's Formula, 

$$\begin{align*}p^2q^2&=(ac+bd)^2-4\left[K^2-(s-a)(s-b)(s-c)(s-d)\right]\tag{7}\end{align*}$$

Isolating $K$ and factorizing you get $(2)$. Still very Bretschneider, isn't it? Coolidge's proof can be found in $[2]$.

References

$[1]$ Andreescu, Titu & Andrica, Dorian, Complex Numbers from A to...Z, Birkhäuser, 2006, pp. 207–209.

$[2]$ Coolidge, J. L. "A Historically Interesting Formula for the Area of a Quadrilateral." Amer. Math. Monthly 46, 345-347, 1939.

Two Identities and their Consequences (draft)

Renowned British mathematician John Conway, in correspondence with Peter Doyle, used two trigonometric formulae to prove Heron's formula. John Casey, in his book "A Treatise of Plane Trigonometry" used two other trigonometric formulae to prove the Brahmagupta's formula. It turns out that the two trigonometric formulae used by Casey for a cyclic quadrilateral generalize the two used by Conway for a triangle. No one seems to have wondered if the two formulae used by Casey could be generalized to a general quadrilateral and use it to prove the Bretschneider's formula. The last link in this chain is precisely what my last article is about!

Article: Two Identities and their Consequences

Abstract. In this note we prove the Heron’s formula (although known, see Conway’s dicussion in $[7]$), the Brahmagupta’s formula (also known, see $[6]$) and the formula for the area of a bicentric quadrilateral (possibly new, see $[12, 13]$), $\sqrt{abcd}$, based on two lesser-known trigonometric formulae $[6, 16]$ involving sine, cosine, the semiperimeter and the side lenghts of a cyclic quadrilateral. Once the two trigonometric formulae have been established (and the necessary adjustments made), the proofs of these area theorems are greatly simplified. Furthermore, we present a generalization of the two aforementioned trigonometric formulae and use it to give an alternative proof of Bretschneider’s formula. Since all these area theorems can be derived from this new generalization, the approach presented in this note, unlike others, provides a more holistic view of these theorems. Our main result for a general convex quadrilateral are the identities

\[ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d)\]

and

\[bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}=(s-b)(s-c),\]

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

Regarding the novelty of the results and proofs presented in this article, I have consulted Martin Josefsson (whom I consider an expert on these issues) and this was the message he sent me:

"Dear Emmanuel,

 

I like your paper, especially how you put these important formulas in a single framwork. I cannot say that I remember seeing the identities (4) and (5) anywhere else before.

 

I have seen (somewhere on the Internet) that proof of Heron's formula using half angle triangle formulas before - but not from your point of view of using cyclic quadrilateral formulas and setting one side = 0. These cyclic quadrilateral half angle formulas are, as you say, not so well known, but both them and your proof of Brahmagupta's formula can be found - more or less in the same way, but with fewer details - in Casey's 1888 book "A Treatise on Plane Trigonometry", see the attachment.

 

About the structure of the article I have not much to say, except perhaps to eliminate the penultimate step in the proofs of Theorems 1 and 5 where you explain how to introduce the semiperimeter in the formulas.

 

Even though much has already been written about these formulas, the ideas for proving Bretschneider' formula and the area of a bicentric quadrilateral are novel as far as I know. I hope you get your paper published.

 

Best regards,

Martin"

The identities $(4)$ and $(5)$ mentioned by Martin are now identities $(5)$ and $(6)$ in the present version of the paper. Martin's work on these issues can be found in the following link:

Martin Josefsson's research work

Below I've added an incomplete concept map of identities $(5)$ and $(6)$ so you can better appreciate the way they relate to other well-known identities.

Update. The article has been published by MATINF. See here.

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.

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

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

$\square$