4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ (I am finally asking this because I am trying for one month, and failing, and this is an "excuse" to myself for not doing olymp. geometry) $\endgroup$ – user441034 Apr 27 '17 at 13:48
  • 3
    $\begingroup$ You're making it more mysterious than it deserves to be. It really is just similarity, as I tried to emphasize. If two ratios are equal, then you can cross-multiply -- as simple as that. $\endgroup$ – quasi Apr 29 '17 at 15:55
  • $\begingroup$ Is there any proof/explanation which clearly demonstrates the invariance of the power of a point ? We can't know what meets your criteria for being a proof/explanation which "clearly demonstrates" blah, blah. If you are going to ask this sort of question, you need to include, for potential answerers, specific and objective criteria that meets your understanding fore an answerthat "clearly demonstrates..." blah, blah. $\endgroup$ – amWhy Jun 1 '17 at 16:38
  • $\begingroup$ The figure on the page en.wikipedia.org/wiki/Tangent-secant_theorem is (in my opinion) a proof without words demonstrating the invariance of the power of a point outside the circle. It's hard to imagine how you get much more of an "aha! moment" than that. Of course, to get the "aha!" you must already be familiar with inscribed angles, and it helps if you already recognize a general pattern in proofs that progress from similar figures to ratios $\dfrac pq=\dfrac st$ to products $pt = qs.$ This is really basic Euclid's-Elements stuff. $\endgroup$ – David K Jun 1 '17 at 20:12
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ -1: This is just a restatement of the exact same proof from Wikipedia, and this is surely not what OP wants. $\endgroup$ – user403711 Apr 28 '17 at 7:31
  • 1
    $\begingroup$ @Arbitrary Kangaroo -- No, not really. The key idea is that there are similar triangles in play. Once you see that, invariance follows easily. If you don't see that, the logic is just a sequence of statements with no clear strategy. $\endgroup$ – quasi Apr 28 '17 at 8:13
  • 2
    $\begingroup$ The similarity is "obvious" if you see it. As I said, it's the key. As to the question of why $\angle CBA = \angle PCA$, they can be regarded as inscribed angles in the same arc. To see it visually, consider the angle $QCA$ where $Q$ is a variable point on minor arc $AC$, very close to $C$. Now slide $Q$ towards $C$. $\endgroup$ – quasi Apr 28 '17 at 9:01
  • 1
    $\begingroup$ The main point of my post is to show the proof in a stripped down way, emphasizing that the invariance formula is just similarity, and nothing more. $\endgroup$ – quasi Apr 28 '17 at 9:05
  • $\begingroup$ See the updated question $\endgroup$ – user441034 Apr 29 '17 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy