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...