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Dotted line is parallel to the diameter; radii are equal length.

Inscribed angle theorem, special case

The upper angle is the inscribed angle; the double angle at the center is the central angle.

  1. The upper triangle is isosceles, because radii are equal; therefore its base angles are equal.

  2. The leftmost angle (with the dotted line) equals the inscribed angle, because they are alternate interior angles.

  3. The double angle on the left equals the central angle, again by alternate interior angles.

Is there a better way to see this?

Note: This special case works whether the central angle is acute (as shown) or obtuse; it also works if reflected to the other side.

  1. The first generalization is to when the arms on either side of the diameter shown is to use this construction on both sides, and add all the angles.

  2. The second generalization is to where the arms are both on the same side of the diameter shown, and to subtract the lesser angles from the greater.

There's more details of the generalizations at Khan Academy's High School Geometry's Inscribed angle theorem proof ( review text; video ), but they use algebra for the special case/lemma, whereas the above is geometrical.

I really liked their reuse of the special case as a lemma - very elegant. But hard to visualize the complete proof of the special case, because it relied on triangle internal angles summing to $180^\circ$. With the great answers to this question, especially @EthanBolker's, I got a way to visualize it - which surprisingly also lead to a simpler proof, by not needing the angle supplementary to the central angle.

(I couldn't find a way to label the parallels/radii with arrows/strike-throughs at geogebra, though an otherwise excellent tool).

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    $\begingroup$ +1. I like your approach. :) As to the question, "Is there a better way to see this?" ... There may not be. Any argument that resorts to the angle-sum theorem is implicitly using the alternate interior angles theorem, while adding an extra step (plus algebra). Some other argument that goes directly from parallelism to the Inscribed Angle Theorem would, I think, be as good as yours, but not necessarily "better". $\endgroup$
    – Blue
    Apr 12, 2018 at 5:42
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    $\begingroup$ In Geogebra, (1) right click the line segment drawn to open the ‘properties’ popup window; (2) if necessary, scroll down to the last option box to include a ‘decoration’ to the line segment drawn. This will not work for rays or lines. $\endgroup$
    – Mick
    Apr 12, 2018 at 8:07
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    $\begingroup$ @hyperpallium: Take a look: trigonography.com/2018/04/12/… $\endgroup$
    – Blue
    Apr 12, 2018 at 9:27
  • $\begingroup$ @Blue Thanks, it's like being written up in a textbook! Being horizontal with labelled points is like seeing it for the first time. I'm unsure of the conventions in geometry, so I defer to your judgement, but: shouldn't segments $\overline{PR}$ and $\overline{OQ}$ be solid? Also, shouldn't the parallel arrows be the same colour and boldness? (Great you have them, unlike mine!) Finally, this may be the "equivalent parallel postulate" and over my head, but I haven't seen equalities being equivalent to parallel lines (as in the line with $\overleftrightarrow{PQ}\parallel\overleftrightarrow{RS}$). $\endgroup$
    – hyperpallium
    Apr 12, 2018 at 11:01
  • $\begingroup$ @Blue Don't worry too much about my comments, it's just feedback; what you've done is the most appreciation I've received in a long time! $\endgroup$
    – hyperpallium
    Apr 12, 2018 at 11:08

2 Answers 2

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Euclid used your figure, but without the parallel, and in the more general form you mention in Note 1, to prove the theorem in question in Elements, III, 20. So you're in good company. He relied on the equal angles in an isosceles triangle, and the fact that the exterior angle of a triangle is equal to the sum of the opposite interior angles, shown in I, 32. The latter used I, 29, which rests on (parallel) Postulate 5, so @Blue's comment is apt. A more direct showing may be hard to find.

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  • $\begingroup$ Would it be fair to say it's simpler than Euclid's? $\endgroup$
    – hyperpallium
    Apr 15, 2018 at 1:51
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    $\begingroup$ If you make explicit the proofs of the two generalizations you mention, how is your argument different from Euclid's? Doesn't it rely on the same truths? $\endgroup$
    – Edward Porcella
    Apr 15, 2018 at 19:08
  • $\begingroup$ Oh, I see what you mean now. Elements I, 32. has a parallel construction very similar to the other answer I mentioned. So inlining it into III, 20. does give the same proof as mine (just as you said), with no duplication. It's striking I, 32 has a lopsided look compared with the other answer - but like an odd piece of lego or bicycle tool, it's just the shape needed here, and according to djoyce, in many other places in Elements too. $\endgroup$
    – hyperpallium
    Apr 16, 2018 at 8:44
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    $\begingroup$ I agree that Euclid $I, 32$ has a lopsided look compared with the figure in the other answer you mention. I believe Aristotle also preferred that one. Please edit in the links--something I need to learn how to do,--and send your diagram. $\endgroup$
    – Edward Porcella
    Apr 16, 2018 at 15:56
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    $\begingroup$ I've edited in the links (they're peer-reviewed, so might take a while). Links are square-backeted text, then parenthesised url: [text](url). Here's the first one: [Elements, III, 20](https://mathcs.clarku.edu/~djoyce/elements/bookIII/propIII20.html) (you can edit your answer to see the others). The "help" link (eg, below "Add Comment") gives a quick reminder without leaving the page. BTW I found the links by googling e.g. "Elements, III, 20", and those pages have links to the relied on postulates, on the right hand side. BTW did you know these postulates off the top of your head? $\endgroup$
    – hyperpallium
    Apr 17, 2018 at 8:52
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This illustration of Edward Porcella's answer integrates I, 32 into the first part of III, 20.

Euclid Elememts III 20

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  • $\begingroup$ I prefer your first (of the four above) diagram as an illustration of the inscribed angle theorem; it's the simplest and least cluttered. It doesn't seem necessary to mark the circle's radii as equal. Why not leave my answer as is, and edit yours down to just that first diagram, which is arguably a pictorial proof of the inscribed angle theorem? $\endgroup$
    – Edward Porcella
    Apr 27, 2018 at 15:35
  • $\begingroup$ @EdwardPorcella BTW in the first (of the four above), the radii are "marked" as equal by being orange. I see it's implied by being radii (and adds clutter), though noticing that is a key part of the proof... and leads to the red/blue angle equality, which are also not marked in 1st figure, so it's uninformative to the uninformed. Another option is like the question's diagram (with all angles the same colour). Yet another is, by redrawing, I could try using "ticks" (?, like segment equality) on the angles. (Though unimportant, the clarity of pictorial proofs have great potential,) $\endgroup$
    – hyperpallium
    Apr 28, 2018 at 3:45

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