Here is a false proof. I will prove that a point inside a circle
(except for the center) is on the circle. In other words, the point is on the
circumference of the circle. Return to my Mathematics pages
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© Copyright 2003, Jim Loy
Here is a false proof. I will prove that a point inside a circle
(except for the center) is on the circle. In other words, the point is on the
circumference of the circle.
Proof: Circle O has a radius r, and point A is inside the circle, but is not the center. Draw line OA. Draw point B, so that (OA)(OB) = r^2 where r^2 means r squared. Bisect AB at C. The perpendicular bisector of AB meets the circle at D and E. These are lengths: OA = OC - AC, OB = OC+CB, OB = OC+AC. Multiplying OA times OB: (OA)(OB) = (OC-AC)(OC+AC) = OC^2-AC^2. By The Pythagorean Theorem, (OA)(OB) = (OD^2-CD^2)-(AD^2-CD^2) = OD^2-AD^2 = r^2-AD^2. We constructed (OA)(OB) to be r^2. So (OA)(OB) = (OA)(OB) - AD^2. So AD^2 = 0, and AD = 0. So A and D are the same point, and A is on the circumference of the circle.
Flaw in the proof: Of course point C is outside the circle, and there are no points D and E.