Draw a circle. Now pick two points (A,B) on the periphery of that circle, so that a line between A and B will have to pass through the center of the circle. The line AB is, in other words, the diameter of the circle. Now, pick a point C anywhere on the periphery of the circle. Prove that in any such triangle ABC, angle C is 90 degrees. -- I have no idea. At first I was even reluctant to agree that C is always 90 degrees, but making random points on a circle seem to confirm this. The best I can come up with is that if I draw a normal from point C to AB, then that normal will always make 90 degree angles, so both "split-triangles" must be similar and so the original ABC must be similar as well, and contain a 90 degree angle. I don't think that constitutes a proof.