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How do I prove that three non-collinear points lie on a circle?
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Any three non-collinear points form the vertices of a triangle.  Let the Vertices be called A, B, and C.
Any point on the perpendicular bisector of side AB is equidistant from A and B. 
Any point on the perpendicular bisector of side AC is equidistant from A and C.
Since the points are not collinear, the perpendicular bisectors intersect.
The point of intersection of the perpendicular bisectors is equidistant from A, and B, and C.
With the point of intersection as the center and its distance from A, or B, or C as the radius, draw a circle.
Find the Circumcenter.
The Circumcenter is the point where the perpendicular bisectors of each side intersect.  So, if you can find all three perpendicular bisectors of any triangle, and extend those lines long enough, you'll notice that all three lines intersect at the same point.  This point is called the Circumcenter.
The cool thing is that the Circumcenter is also the center of a circle that goes through all three vertices of the triangle formed by the three non-collinear points.  This means that any three non-collinear points do in fact lie on a circle.