General proof of the Power of a Point Theorem (uncomplicated) The definition of The Power of a Point Theorem in the Wolfram Demonstrations Project is the following:

If two chords $AB$ and $CD$ in a circle intersect the point $P$, then $|AP||PB|=|CP||PD|$

So far, all the proofs of this theorem I have found are restricted to cases: $P$ is either necessarily inside or outside the circle. What would be an uncomplicated proof of this theorem comprising both cases at once?

In the most-nuanced interpretation, the "power" of a point $P$ with respect to a circle $\gamma$ is a signed value:
$$\operatorname{pow}_\gamma (P) = p^2-r^2 \tag{1}$$
where $r$ is the circle's radius, and $p$ is the distance from $P$ to the circle's center. This power value is positive for $P$ outside $\gamma$ (ie, $p>r$), negative for $P$ inside ($p<r$), and zero for $P$ on the circumference ($p=r$). This trichotomy typically invites breaking the proof that "the power equals product of distances from $P$ to collinear intersections with $\gamma$" into cases, to avoid introducing "signed distances" into the argument.

Coordinate geometry doesn't discriminate against signed distances, so let's take a look at a quick argument there. Let $\gamma$ be centered at the origin, so that its equation is 
$$x^2+y^2=r^2 \tag{2}$$ 
Let $P$ have coordinates $(p,0)$, and suppose a line through $P$ makes an angle $\theta$ with the $x$; we can describe the points on this line by
$$P + t\,(\,\cos\theta, \sin\theta\,) \tag{3}$$
where $t$ amounts to the signed distance along that line $P$. (Points in one direction along the line have positive distances; points in the other direction have negative distances.) To find the points where the line meets the circle, we merely substitute the coordinates from $(3)$ into $(2)$.
$$(p+t\cos\theta)^2 + (0+t\sin\theta)^2 = r^2 \qquad\to\qquad t^2 + 2 p t \cos\theta + p^2-r^2 = 0 \tag{4}$$
By Vieta's formula (or by solving the quadratic and multiplying), we find that the product of the roots of $(4)$ is the constant term. This says exactly that 

The product of the signed distances from $P$ to collinear points of intersection with $\gamma$ is $p^2-r^2$ (which is to say, the power of $P$ with respect to $\gamma$).

The reader should note that, if $P$ is inside the circle, then $P$ is between the points of intersection; thus, those points have oppositely-signed distances from $P$, making the product of those distances negative. Contrariwise, if $P$ is outside the circle, the points have same-signed distances, making the product positive. Finally, for $P$ on the circle, at least one of those distances is $0$, making the product $0$, too.
