Talk:Circle is Bisected by Diameter

Proof 1 is flawed. The following claim is unsupported:

"Thus it will be possible to find a diameter DE passing through C such that DC≠CE."

To fix it, we could use "vertical angles are congruent" and "SAS" twice apiece, but this presumes that SAS is an axiom, apart from being unnecessary.

I have given an elementary proof which I believe to be appropriately simple and correct.


 * So all that needs to be done is for those to be invoked. Does SAS depend on this? If so then indeed it is flawed, but without studying it in detail it does not appear so.


 * The point is, there is no reason why we should not keep both proofs. --prime mover (talk) 13:08, 8 November 2016 (EST)


 * SAS does not depend on this, but the proof doesn't even mention vertical angles or SAS. The statement I quoted pops out of nowhere (if it follows for you, please tell me). I see no way to repair it without making it a more complicated version of the proof I gave.


 * Please sign your posts, by the way. No, I agree, it does not mention vertical angles or SAS, that was you explaining how it could be fixed. And indeed, by invoking those propositions, the existing proof would indeed be complete. A more complicated version of yours? Maybe so, but your assertion that arc $AOB$ equals arc $BOA$ is itself assumed without proof. --prime mover (talk) 13:55, 8 November 2016 (EST)


 * Ok, thanks for telling me. I believe arcs are congruent, by definition, when their central angles are congruent. I have updated my proof.--AuSmith (talk) 16:42, 8 November 2016 (EST)


 * Well actually, what I was thinking of was that there's a proposition in somewhere which specifically proves it, which you may care to use if you wish. --prime mover (talk) 23:07, 8 November 2016 (EST)


 * The best I can determine from Book III is that Euclid only compares angles that segments and sectors make and never discusses arcs. According to how geometry is currently taught, it is the definition of the measure of an arc. I did run across an excerpt, regarding the proof that diameters bisect circles, in Heath's translation of Book I: "Proclus gives as the reason of the property "the undeviating course of the straight line through the centre" (a simple appeal to symmetry), but adds that, if it is drawn and one part of the circle applied to the other; it is then clear that they must coincide, for, if they did not, and one fell inside or outside the other, the straight lines from the centre to the circumference would not all be equal: which is absurd."--AuSmith (talk) 09:41, 9 November 2016 (EST)


 * Just that the whole of is here on, some nutcase with a sense of the absurd obviously thought it would be funny. --prime mover (talk) 13:59, 9 November 2016 (EST)


 * It is quite the resource! Thanks--AuSmith (talk) 16:02, 9 November 2016 (EST)


 * Aha! Found it: Straight Lines Cut Off Equal Arcs in Equal Circles. --prime mover (talk) 16:24, 9 November 2016 (EST)


 * You mean this: Equal Angles in Equal Circles. Thanks - I will include it. Back to the main point, I still do not see how to resolve the gap in proof 1.--AuSmith (talk) 16:50, 9 November 2016 (EST)