Power of a point is a famous proof in elementary geometry, which states that:
For a circle $\omega$ and a pont $P$ outside it, for any line through $P$ which intersects $\omega$ at $A$ and $B$, the quantity $PA.PB$ is invariant for any $A$, $B$.
I can find the proof in wikipedia, and in all other places, but I am not finding the proof intuitive enough to see the invariance of the multiplication at a glance, and have an "Aha! Gotcha" insight.
Don't get me wrong, I understand the proof perfectly; But it's like the well mentioned sum $\sum_{i=0}^{n}n^3 = [\sum_{i=0}^{n}n]^2 $ - you can prove that it's correct, but you can't see it at a glance. (For that particular equation, some Proof-without-words gives the proper intuition).
Is there any proof/explanation which clearly demonstrates the invariance of the power of a point ?
Update: The other answer (which doesn't provides any new insight, but) poses another good spinoff question, for circle $\omega$, and arc ${AB}$ and any point $P$ on $\omega$, $\angle {APB}$ is constant. Same situation as above, I know the proof, but I don't have any a-glance insight on its truth.
Draw a tangent line from $P$ to $\omega$, meeting $\omega$ at $C$, say.
Draw chord $AC$.
Now find similar triangles . . .
Angles $CBA$ and $ACP$ are inscribed in the same arc, hence
\begin{align*} &\angle CBA = \angle ACP\\[4pt] \implies\;&\angle CBP = \angle ACP \end{align*}
Since triangles $CBP$ and $ACP$ also have a shared angle at vertex $P$, it follows that the triangles are similar.
Then, letting $k = PC$, similarity yields \begin{align*} &\frac{PB}{PC}=\frac{PC}{PA}\\[6pt] \implies\;&(PA)(PB) = k^2 \end{align*}
Thus, the value of the product $(PA)(PB)$ is constant, independent of the choice of secant.