Euclid's Proposition 6 Book I proof by circle

Asked Nov 05 '16 at 18:59

Active Nov 05 '16 at 19:22

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Euclides proves proposition 6 in book I using a reductio ad absurdum proof assuming that line AB is less than line AC.

Couldn't we just draw a circle with center A and distance B, and by definition 15 prove that AB = AC, as described in the following figure?

edited Nov 05 '16 at 19:22

asked Nov 05 '16 at 18:59

felipegf

Could you give the complete proof that you are proposing? – Eric M. Schmidt Nov 05 '16 at 19:06
Yes, one moment. – felipegf Nov 05 '16 at 19:09
Ok, there it is. – felipegf Nov 05 '16 at 19:21
But you need show that if a triangle have two angles equals THEN the opposite sides are equal.
If you use a circle for show $AB$ is equal to $AC$ you not saying nothing about the angles.
– sango Nov 05 '16 at 19:22
So, what exactly is my proof stating? That this particular triangle has sides AB = AC? – felipegf Nov 05 '16 at 19:28
Then it seems to me that I don't know yet how to build mathematical proofs. I should use that information about the angles and build upon that to arrive at my conclusion? – felipegf Nov 05 '16 at 19:30
You are trying prove, if a triangle have to side equals then the opposite angles are equals. – sango Nov 05 '16 at 19:33
@felipegf The proposition is with the form $p \implies q$, but Euclides use the equivalence $\lnot q \implies \lnot p$ – sango Nov 05 '16 at 19:35

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