The Butterfly Theorem

Let be the midpoint of the chord of a circle. Let and be two other chords of the circle passing through and let and be the intersection points of with and respectively. Then is the midpoint of the segment . Proof: The space of all degree two polynomials vanishing at the four points and is of dimension 2*. Thus it is spanned by the equation of the circle and the equation of the union of the lines and . Choose coordinates such that the line is the -axis and the point is the origin. Then the coefficient of in both our spanning polynomials is equal to zero, hence this coefficient is also zero in the equation of the union of and . Therefore is the midpoint of the segment , as required.

*To see this there is a six dimensional space of degree two polynomials and we are imposing one linear condition each time we require the polynomial to vanish at a point. These four conditions are linearly independent since it is easy to find degree two curves passing through three of these points and not the fourth (e.g. a union of two lines).