# Butterfly Theorem Proved by Projection

Hubert Shutrick
August 2012

This theorem is stated and proved elsewhere.

The horizontal strut of the kite is a segment PQ of the x-axis with midpoint the origin O and its vertical strut is a segment RS of the y-axis with R above and S below O . The line AOB has A on PR and B on QS and the line COD has C on RQ and D on PS . The lines AD and BC intersect PQ at X and Y . The theorem states that XO = OY .

It can be changed into an incidence theorem by reflecting the right-hand side into the left-hand side through the point O as illustrated. It is now required to show that AD, PO, B'C' concur.

For fixed AO , consider the projectivity that takes X to Y' as follows: project PO onto PS by the point A ; project PS to PR' by O ; project PR' to PO by B' .

In the special case when X is chosen such that DC' is the reflection of B'A in PO it is obvious that it goes to itself, if X=O , then it goes to points along the line AO and back to O and, if X=P , then it just remains there for each projection. However, a projectivity is uniquely determined by images of three points so this one is the identity and Y'=X .

The same method can be used to prove the theorem directly, without the reflection in O. For fixed AB, define a projectivity by three projections that takes X to Y and note that it leaves O fixed and takes P to Q and, if X is the point so that the diagram is the symmetrical about RS, then OX = OY. Since the projectivity agrees with the involution that reflects in RS for three positions of X, it must coincide with the involution.