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Show that the affirmation

"For every three points (not on the same line) on the plane, one can find a circle that contains these three points"

is equivalent to

"Given a line and a point out of this line, there is only one line passing through this point and parallel to the first one".

I was able to show that the second affirmation implies the first one:

Consider a triangle ABC. Let M be the midpoint of AB. Mark P such that the angle PMA = 90° = PBM. Now, let N be the midpoint of BC. Using the second statement, there is only one line passing through N parallel to MP. This line can't be orthogonal to BC, otherwise we would have a triangle with two rect angles. So, the line passing through N perpendicular to BC meets PM in a point O. Now, it's easy to construct such a circle.

But I couldn't prove the reciprocal...

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    Well, R. Hartshorne, Geometry: Euclid and Beyond, exercise 33.11 on pages 302-303. He calls the circle thing Bolyai's Axiom, in this case Farkas Bolyai. He gives a pretty complete recipe. So it depends on what background you have set up when you get to the question. Marvin mostly wants you to think about it. I'm in Marvin's book, mostly pages 524-526.2012-05-05

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