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I recently watched Douglas Hofstadter recall at the beginning of a talk about Feuerbach's Theorem the fact that under circle inversion, circles (which do not pass through the center of inversion) map to circles. He claimed it used the fact that if two inscribed angles cut the same arc, they must have equal measure (chord-angle theorem).

I tried to rediscover the proof, but concluded that his chord-angle theorem was useless. Assuming his definition of circle inversion indeed inverts distances from the center, he must also use some result about distance, but power of a point proves circle inversion easily without direct reference to the chord-angle theorem.

Is there a well known proof involving the chord-angle theorem?

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    $\begingroup$ It may help to know that what you call the "chord-angle theorem" is typically called the "Inscribed Angle Theorem". Perhaps a web search using that term will be fruitful. $\endgroup$
    – Blue
    Sep 11, 2017 at 22:33
  • $\begingroup$ Ah yes, thank you. It's one of my favorite theorems, yet I sometimes have trouble remembering the name. I googled some and keep coming across an argument made with the more rudimentary Thales theorem (diameter subtends a right angle) with some angle chasing. I think it could be polished enough to be a pleasing proof.I guess I was expecting something shocking from the way Hofstadter talked about it. $\endgroup$ Sep 12, 2017 at 0:02
  • $\begingroup$ I suppose after some more consideration that the Power of a Point theorem is circle inversion. And, there is a marvelous proof of PoP which uses the Included Angle Theorem. Perhaps you can think of them together as a marvelous proof that circles invert to circles. cut-the-knot.org/pythagoras/PPower.shtml $\endgroup$ Sep 12, 2017 at 1:28

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Here's a straightforward proof of the circles-invert-to-circles theorem, where the linchpin is the Inscribed Angle Theorem (more-specifically, Thales' Theorem: An angle inscribed in a semicircle is a right angle.).

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Consider a point $C$ on some circle. We wish to invert $C$ in a circle with center $O$. Let $\overline{AB}$ be the diameter of $C$'s circle such that $O$, $A$, $B$ are collinear. Let the inversions of $A$, $B$, $C$ be $A^\prime$, $B^\prime$, $C^\prime$.

Recall that the nature of inversion is that $$|\overline{OX}||\overline{OX^\prime}| = \left(\;\text{radius of $\bigcirc O$}\;\right)^2 \tag{1}$$ The radius is irrelevant; what matters is the constant product, so that we have $$|\overline{OA}||\overline{OA^\prime}| = |\overline{OB}||\overline{OB^\prime}| = |\overline{OC}||\overline{OC^\prime}| \tag{2}$$

This allows us to argue as follows: $$\begin{align} \frac{|\overline{OA}|}{|\overline{OC}|} = \frac{|\overline{OC^\prime}|}{|\overline{OA^\prime}|} &\implies \triangle OAC \sim \triangle OC^\prime A^\prime \implies \angle A^\prime C^\prime C = \angle A \tag{3a} \\[8pt] \frac{|\overline{OB}|}{|\overline{OC}|} = \frac{|\overline{OC^\prime}|}{|\overline{OB^\prime}|} &\implies \triangle OBC \sim \triangle OC^\prime B^\prime \implies \angle B^\prime C^\prime O = \angle B \tag{3b} \end{align}$$

Consequently, $$\angle ACB \cong \angle A^\prime C^\prime B^\prime \tag{4}$$

Now, Thales both sets us up and brings us home ...

$$\begin{align} \text{$C$ lies on a semicircle on $\overline{AB}$}\quad &\implies\quad \text{$\angle ACB$ is a right angle} \\[6pt] &\implies\quad \text{$\angle A^\prime C^\prime B^\prime$ is a right angle} \\ &\implies\quad \text{$C^\prime$ lies on a semicircle on $\overline{A^\prime B^\prime}$} \end{align}$$

Thus, the points on a circle invert to points on another circle, and (with some handwaving about continuity and bijection) we're done. $\square$


In point of fact, the restriction to Thales' Theorem and right angles and diameters is unnecessary. We can take $\overline{AB}$ to be any (non-degenerate) chord of $C$'s circle, so long as it's collinear with $O$, and the proof effectively works as-is, requiring only that we replace "right angle" and "semi-circle" with appropriately-generic descriptors.


By request, here are diagrams for varying configurations. In all cases, $(2)$ leads to similar triangles that imply $(3a)$, $(3b)$, and, ultimately $(3c)$, so that $\angle A^\prime C^\prime B^\prime$ is a right angle. "Usually", this means that $C^\prime$ lies on a regular circle with diameter $\overline{A^\prime B^\prime}$; when $A=O$, we have that $A^\prime$ is the "point at infinity", and we see that $C^\prime$ lies on the perpendicular line (that is, a circle of infinite radius) through $B^\prime$.

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  • $\begingroup$ Sorry for reviving this old answered post, but what does it look like to use these methods to prove all possible cases (circle through the center of inversion circle, circle crossing the inversion circle, circle encompassing the center of the inversion circle, etc.) ? Could you link a good source? Most sources I find include a discussion of this that goes like "We will take it on faith that circles invert to circles, because there are many cases to consider" :( As always, thanks blue! You're amazing! $\endgroup$ Dec 20, 2018 at 19:27
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    $\begingroup$ @MathEnthusiast: See my edit. $\endgroup$
    – Blue
    Dec 20, 2018 at 20:55
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    $\begingroup$ Spectacular! Thanks so much! $\endgroup$ Dec 21, 2018 at 0:02
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    $\begingroup$ @MathEnthusiast: In the $A=O$ case, we want that $C^\prime$ lies on the perp at $B^\prime$; this actually follows from the similarity in $(3b)$, which guarantees a right angle at $B^\prime$. That said, it's not wrong —merely inefficient — to argue as in the other cases. Taking "$\angle A^\prime C^\prime C$" to mean the appropriate angle between $\overline{CC^\prime}$ and the "horizontal" through $C^\prime$, the angle equality in $(3a)$ holds by parallelism; with $(3b)$, this implies $(4)$, giving a right angle at $C^\prime$. This makes $\overline{B^\prime C^\prime}$ "vertical", as desired. $\endgroup$
    – Blue
    Dec 22, 2018 at 23:21
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    $\begingroup$ Aha! I missed the similarity in 3b that showed OB'C' was a right angle, thanks for the clarification $\endgroup$ Dec 23, 2018 at 0:21

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