Need help for geometry problem ($\frac {\Delta_0}{\Delta}= |PO^2- R^2|$) I recently discovered some property of triangles when I was trying to find a solution to some apparantly unrelated problem, and the path that I took to get there was a long and complicated one. And yet I now have a simple innocent looking property that I did technically 'prove' but I cannot prove it as a standalone problem without going back to the long and complicated path. I need your help to get a reasonable proof. Here's the problem:
Let $ABC$ be a triangle with sides $a,b,c$ and a point $P$ in it's plane. $O$ is the circumcentre, $R$ is the circumradius and $\Delta$ is the area of $ABC$. Let $PA =x, PB=y, PC=z$. For algebraic convenience, we denote $\Delta_0$ to be the 'area' of the 'triangle' that would have sides of "$ax,by,cz$"
Prove the following:
$$\frac {\Delta_0}{\Delta}= |PO^2- R^2|$$
What gives me confidence that this is a correct theorem is the fact that the cases $PO =R$ and $PO=0$ are easy to verify. But I am unable to prove the general case. I hope a reasonable proof exists..
 A: There's an interesting generalization of this result to three dimensions.
Take $\triangle ABC$ (with angles $\alpha$, $\beta$, $\gamma$ and circumradius $r$) to lie the $xy$-plane using coordinates
$$\begin{align}
A &= (r\cos\theta,r\sin\theta,0)\\
B &=(r\cos(\theta+2\gamma),r\sin(\theta+2\gamma),0)\\
C &=(r\cos(\theta-2\beta),r\sin(\theta-2\beta),0)
\end{align}$$
Let point $P$ lie in the $xz$-plane, with $p:=|OP|$ and $\phi$ the angle between $\overline{OP}$ and the positive $x$-axis; thus,
$$P = (p\cos\phi,0,p\sin\phi)$$
Then we have
$$\begin{align}
u^2 &:= |BC|^2|PA|^2 = 4 r^2 \sin^2\alpha\cdot\left(p^2 + r^2 - 2 p r \cos\phi\cos\theta\right) \\
v^2 &:= |CA|^2|PB|^2 = 4r^2\sin^2\beta\cdot\left(p^2 + r^2 - 2 p r \cos\phi\cos(\theta+2\gamma)\right) \\
w^2 &:= |AB|^2|PC|^2 = 4r^2\sin^2\gamma\cdot\left(p^2 + r^2 - 2 p r \cos\phi\cos(\theta-2\beta)\right)
\end{align}$$
By Heron's Formula, 
$$\begin{align}
\Delta_0^2 &= \frac1{16}(u+v+w)(-u+v+w)(u-v+w)(u+v-w) \\[4pt]
&= \frac1{16}\left(-u^4-v^4-w^4+2u^2v^2+2u^2w^2+2v^2w^2\right) \\[4pt]
&= \text{... Mathematica ...} \\[4pt]
&= 4 r^4 \sin^2\alpha\sin^2\beta\sin^2\gamma \left(p^4 + r^4 - 2 p^2 r^2 \cos 2\phi\right) \\[4pt]
&=\left(\frac12\cdot 2r\sin\alpha\cdot 2r\sin\beta\cdot\sin\gamma\right)^2\left(p^4+r^4-2p^2r^2(2\cos^2\phi-1)\right)\\[4pt]
&=|\triangle ABC|^2\left(p^2 + r^2 - 2 p r \cos\phi\right) \left(p^2 + r^2 + 2 p r \cos\phi\right)
\end{align}$$

Interestingly, if we define $R$ and $R'$ as the points where the $x$-axis meets the circumcircle —that is, the points where the plane through $\overline{OP}$, perpendicular to the plane of $\triangle ABC$, meets the circumcircle— the above becomes

$$\Delta_0 = |\triangle ABC| \;|PR|\;|PR'| \tag{$\star$}$$

For $P$ in the plane of $\triangle ABC$, the reader may recognize the product $|PR||PR'|$ as the absolute value of the power of $P$ with respect to the circumcircle. That product is equal to $\left|p^2-r^2\right|$, yielding the result in the question. $\square$
It seems as though $(\star)$ is trying to tell us something, but I'm not sure what it is ...
A: 
The property checks out 
with
a straightforward proof using 
complex numbers representation, with
the circumcenter at the origin,
\begin{align}
O&=0
,\\
A&=R
,\\
B&=R\,\exp(\mathbf i\,\phi);
,\\
C&=R\,\exp(\mathbf i\,\psi);
,\\
P&=|PO|\,\exp(\mathbf i\,\theta)
.
\end{align} 
\begin{align} 
a^2&=(B-C)\overline{(B-C)}
,\\
b^2&=(A-C)\overline{(A-C)}
,\\
c^2&=(A-C)\overline{(B-A)}
,\\
x^2=|PA|^2&=(P-A)\overline{(P-A)}
,\\
y^2=|PB|^2&=(P-B)\overline{(P-B)}
,\\
z^2=|PC|^2&=(P-C)\overline{(P-C)}
.
\end{align}
where $\overline{z}$ is the complex conjugate of $z$.
The square of the areas is calculated using a variant of the  Heron’s formula:
\begin{align} 
S^2&=
\tfrac1{16}\,(4a^2b^2-(a^2+b^2-c^2)^2)
,\\
S_0^2&=
\tfrac1{16}\,(4x^2a^2y^2b^2-(x^2a^2+y^2b^2-z^2c^2)^2)
,\\
\frac{S_0^2}{S^2}
&=(R^2-|PO|^2)^2
.
\end{align}
Corresponding Maxima session:
Maxima 5.38.1 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.12
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) _(z):=conjugate(z)$
(%i2) fS2(a2,b2,c2):=1/16*(4*a2*b2-(a2+b2-c2)^2)$
(%i3) declare([A,B,C,P,O],complex)$
(%i4) declare([R,PO,phi,psi,theta,a2,b2,c2,PA2,PB2,PC2,S2,S02],real)$
(%i5) A:R$
(%i6) B:R*exp(%i*phi)$
(%i7) C:R*exp(%i*psi)$
(%i8) P:PO*exp(%i*theta)$
(%i9) a2:(B-C)*_(B-C)$
(%i10) b2:(A-C)*_(A-C)$
(%i11) c2:(B-A)*_(B-A)$
(%i12) PA2:(P-A)*_(P-A)$
(%i13) PB2:(P-B)*_(P-B)$
(%i14) PC2:(P-C)*_(P-C)$
(%i15) S2:fS2(a2,b2,c2)$
(%i16) S02:fS2(a2*PA2,b2*PB2,c2*PC2)$
(%i17) factor(S02/S2);
                                      2         2
(%o17)                        (R - PO)  (R + PO)

