Butterfly Theorem

Revision as of 12:32, 26 January 2015 by WoweeG (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $XY$.

528px-Butterfly theorem.svg.png

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. $\blacksquare$.

Related Reading

http://agutie.homestead.com/FiLEs/GeometryButterfly.html

http://www.mathematik.uni-muenchen.de/~fritsch/butterfly.pdf

See also

Midpoint

Invalid username
Login to AoPS