Proof - Area of a cyclic quadrilateral So we have a cyclic quadrilateral, as depicted below: 

I have a conjecture that the area of this cyclic quadrilateral equals $$ \dfrac{\sqrt{(a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a)}}{4} $$
I want to prove this. I know that the area of triangle ABC equals $\dfrac{1}{2}ab\sin(B)$ and the area of triangle ACD equals $\dfrac{1}{2}cd\sin(D)$. Seeing as $\sin(B) = \sin(D)$, I figured that the area of the quadrilateral equals $\dfrac{1}{2}(ab+cd)\sin(B)$. 
But I'm stuck here. I have no clue how to get to the result I want. Can anyone provide an insight (and tell me whether or not my work up until now is valid)?
 A: 
The Inscribed Angle Theorem shows that opposite angles of a cyclic quadrilateral are supplementary. Thus,
$$
D=\pi-B
$$
Therefore, $\cos(D)=-\cos(B)$ and $\sin(D)=\sin(B)$.
The Law of Cosines says
$$
e^2=\overbrace{a^2+b^2-2ab\cos(B)}^{\text{as a side of $\triangle ABC$}}=\overbrace{c^2+d^2+2cd\cos(B)}^{\text{as a side of $\triangle ADC$}}
$$
Solving for $\cos(B)$, we get
$$
\cos(B)=\frac12\frac{a^2+b^2-c^2-d^2}{ab+cd}
$$
Furthermore,
$$
\begin{align}
\frac14\sin^2(B)
&=\frac{(2ab+2cd)^2-(a^2+b^2-c^2-d^2)^2}{16(ab+cd)^2}\\
&=\frac{\left((c+d)^2-(a-b)^2\right)\left((a+b)^2-(c-d)^2\right)}{16(ab+cd)^2}\\
&=\frac{((c+d+a-b)(c+d+b-a))((a+b+c-d)(a+b+d-c))}{16(ab+cd)^2}\\
&=\frac{(s-a)(s-b)(s-c)(s-d)}{(ab+cd)^2}
\end{align}
$$
where $s=\frac{a+b+c+d}2$ is the semi-perimeter.
The area of ${\large\unicode{x23E5}}ABCD\ \ $ is the sum of the area of $\triangle ABC=\frac12ab\sin(B)$ and the area of $\triangle ADC=\frac12cd\sin(D)$
$$
\frac12(ab+cd)\sin(B)=\sqrt{(s-a)(s-b)(s-c)(s-d)}
$$
which, as Christian Blatter mentions, is called Brahmagupta's Formula.
A: These might be useful:

Heron's formula:
  $$\large \Delta=\sqrt{s(s-a)(s-b)(s-c)}$$

${}$

Ptolemy's theorem:
  $$\large AC.BD=ac+bd$$

A: This is Brahmagupta's formula; see here:
http://en.wikipedia.org/wiki/Brahmagupta%27s_formula
