Heron's formula vs 'complex cross products' for triangle area? I would like to prove the equivalence between the following two ways of calculating the area of the triangle between points $\;p\;$, $\;q\;$, and $\;r\:$ in the complex plane.
First, there is Heron's formula, $$
\tag{0} \sqrt{s(s-a)(s-b)(s-c)}
$$ where $$
a = \left| p-q \right| \\
b = \left| q-r \right| \\
c = \left| r-p \right| \\
s = {1 \over 2} (a+b+c)
$$
Second, there is $$
\tag{1} \left| {1 \over 2i} (p \times q + q \times r + r \times p) \right|
$$ where $$
x \times y \;=\; {1 \over 2} (\overline{x}y - x\overline{y})
$$ is the 'complex cross product' of complex numbers $\;x,y\;$.  (See my question Complex integral to determine area inside of parameterized closed curve for the reference that gave me $(1)$, by adding the 'directed area' of triangles $\;\triangle 0 p q\;$, $\;\triangle 0 q r\;$, and $\;\triangle 0 r p\;$.)
How do I prove $(0)$ and $(1)$ are equivalent?  It seems it should be a fairly simple calculation, but I've tried multiple approaches and cannot yet get anywhere...
 A: Disclaimer : I will not try here to connect your formulas (0) and (1) 
(see Edit below for connections between Heron's formula and complex numbers).
I will just show here that (1) can be considered in a very plain way.
Indeed, (1) is a direct consequence of the Laplace expansion of the following determinant :
$$A=\frac{i}{4}\begin{vmatrix}p&q&r\\\overline{p}&\overline{q}&\overline{r}\\1&1&1\end{vmatrix}\tag{I}$$
with respect to its last row.
(I) can be obtained from the equivalent rather classical formula (16) in this reference with real coordinates :
$$A=\frac12\begin{vmatrix}x_1&x_2&x_3\\y_1&y_2&y_3\\1&1&1\end{vmatrix}\tag{II}$$
Proof : (II) is related to (I) through the following identity (taking determinants on both sides) :
$$\begin{pmatrix}p&q&r\\\overline{p}&\overline{q}&\overline{r}\\1&1&1\end{pmatrix}=\begin{pmatrix}1& \ \ \ i&0\\1&-i&0\\0& \ \ \ 0&1\end{pmatrix}\begin{pmatrix}x_p&x_q&x_r\\y_p&y_q&y_r\\1&1&1\end{pmatrix}.$$
Edit : 
1) Have a look to this article about Cayley-Menger determinant. It is worth seeing two paragraphs : paragraph $4$ about the fact that Heron's formula can be expressed as a  $4 \times 4$ (Cayley-Menger) determinant, and paragraph $2$ giving a nice proof of Heron's formula using complex numbers.
2) See p. 106 of the following book "Complex Numbers from A to... Z", 2nd Ed., T. Andreescu and D. Andrica, Birkhäuser 2010, for your formula (1) and other formulas.
3) For applications of formula (I), see this interesting article (formula (I) is their formula (4)) and extensions by the same author.
4) Please note that (I) represents the oriented area of the triangle.
5) Formula (1) can be used to express the '$(z,\overline{z})$' equation of the line $[p,q]$ under the form :
$$\begin{vmatrix}p&q&z\\\overline{p}&\overline{q}&\overline{z}\\1&1&1\end{vmatrix}=0$$
(indeed $z$ is aligned with $p,q$ iff the area of triangle $p,q,z$ is zero.)
Thanks to @achille hui for a sign correction in formula (I). 
A: One way is to show that both these numbers are equal to
$$S=\frac{1}{2}|p-r||p-q|\xi(p-r,p-q)$$
where $\xi(u,v)$ is the sine of the angle between vectors $u,v$ i.e.
 $$\xi(u,v)=\sqrt{1-\frac{\langle u,v\rangle^2}{|u|^2\cdot|v|^2}}$$
First, observe that
$$\langle p-r,p-q\rangle=|p|^2-\langle r,p\rangle-\langle p,q\rangle+\langle r,q\rangle=|p|^2-\frac{1}{2}(-|p-r|^2+|p|^2+|r|^2)-\frac{1}{2}(-|p-q|^2+|p|^2+|q|^2)+\frac{1}{2}(-|r-q|^2+|r|^2+|q|^2)=\frac{1}{2}(|p-r|^2+|p-q|^2-|r-q|^2)=\frac{1}{2}(c^2+a^2-b^2)$$
(Btw it is known as cosine law).
Then,
$$S=\frac{1}{2}|p-r||p-q|\sqrt{1-\frac{\langle p-r,p-q\rangle^2}{|p-r|^2\cdot|p-q|^2}}=\frac{1}{2}\sqrt{c^2a^2-\left(\frac{1}{2}(c^2+a^2-b^2)\right)^2}=\frac{1}{2}\sqrt{\frac{1}{4}(2ac-a^2-c^2+b^2)(2ac+a^2+c^2-b^2)}=\frac{1}{4}\sqrt{(b^2-(a-c)^2)((a+c)^2-b^2)}=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$
On the other hand,
$$|\frac{1}{2}(p\times q+q\times r+r\times p)|=\frac{1}{2}|(p-q)\times (p-r)|$$
I will show that for all $u,v$:
$$|u\times v|=|u||v|\xi(u,v)$$
Indeed,
$$|u\times v|^2=|\frac{1}{2}(\overline{u}v-u\overline{v})|^2=\frac{1}{4}\left(|\overline{u}v|^2+|u\overline{v}|^2-2\langle \overline{u}v,u\overline{v}\rangle \right)=\frac{1}{4}\left(2|u|^2|v|^2-2\left(2\langle u, v\rangle^2-|u|^2|v|^2\right)\right)=|u|^2|v|^2-\langle u,v\rangle^2=(|u||v|\xi(u,v))^2$$
The property
$$\langle \overline{u}v,u\overline{v}\rangle=2\langle u, v\rangle^2-|u|^2|v|^2$$
used above can be proved by writing out real and imaginary parts of $u$ and $v$.
