# Difference between revisions of "Butterfly Theorem"

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− | Let <math> | + | Let <math>M</math> be the [[midpoint]] of [[chord]] <math>PQ</math> of a [[circle]], through which two other chords <math>AB</math> and <math>CD</math> are drawn. <math>AD</math> and <math>BC</math> intersect chord <math>PQ</math> at <math>X</math> and <math>Y</math>, respectively. The '''Butterfly Theorem''' states that <math>M</math> is the midpoint of <math>XY</math>. |

− | + | [[File:528px-Butterfly_theorem.svg.png ]] | |

==Proof== | ==Proof== | ||

− | + | This simple proof uses [[projective geometry]]. | |

− | + | First we note that <math>(AP, AB; AD, AQ) = (CP, CB; CD, CQ).</math> | |

+ | Therefore, | ||

+ | <cmath>\frac{(PX)(MQ)}{(PQ)(MX)} = \frac{(PM)(YQ)}{(PQ)(YM)}.</cmath> | ||

+ | Since <math>MQ = PM</math>, | ||

+ | <cmath>\frac{MX}{YM} = \frac{XP}{QY}.</cmath> | ||

+ | Moreover, | ||

+ | <cmath>\frac{MX + PX}{YM + QY} = 1,</cmath> | ||

+ | so <math>MX = YM,</math> as desired. | ||

+ | <math>\blacksquare</math>. | ||

+ | ==Related Reading== | ||

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

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

==See also== | ==See also== | ||

[[Category:Geometry]] | [[Category:Geometry]] | ||

[[Category:Theorems]] | [[Category:Theorems]] | ||

+ | [[Midpoint]] |

## Latest revision as of 12:32, 26 January 2015

Let be the midpoint of chord of a circle, through which two other chords and are drawn. and intersect chord at and , respectively. The **Butterfly Theorem** states that is the midpoint of .

## Proof

This simple proof uses projective geometry.

First we note that Therefore, Since , Moreover, so as desired. .

## Related Reading

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

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