# Difference between revisions of "Butterfly Theorem"

Let $M$ be the midpoint of chord $PQ$ of a circle, through which two other chords $AB$ and $CD$ are drawn. $AD$ and $BC$ intersect chord $PQ$ at $X$ and $Y$, respectively. The Butterfly Theorem states that $M$ is the midpoint of $XY$.

## Proof

This simple proof uses projective geometry.

First we note that $(AP, AB; AD, AQ) = (CP, CB; CD, CQ).$ Therefore, $$\frac{(PX)(MQ)}{(PQ)(MX)} = \frac{(PM)(YQ)}{(PQ)(YM)}.$$ Since $MQ = PM$, $$\frac{MX}{YM} = \frac{XP}{QY}.$$ Moreover, $$\frac{MX + PX}{YM + QY} = 1,$$ so $MX = YM,$ as desired. $\blacksquare$.