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)