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Difference between revisions of "Butterfly Theorem"
 
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Let <math>D</math> be the [[midpoint]] of [[chord]] <math>BC</math> of a [[circle]], through which two other chords <math>EH</math> and <math>FG</math> are drawn. <math>EG</math> and <math>HF</math> intersect chord <math>BC</math> at <math>I</math> and <math>J</math>, respectively. The '''Butterfly Theorem''' states that <math>D</math> is the midpoint of <math>IJ</math>.
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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>.
  
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<geogebra>ac0eaced14d78b0f4ff370ae453962b4db309b5f</geogebra>
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[[File:528px-Butterfly_theorem.svg.png‎ ]]
  
 
==Proof==
 
==Proof==
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{{solution}}
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This simple proof uses [[projective geometry]].
  
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Link to a good proof :
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First we note that <math>(AP, AB; AD, AQ) = (CP, CB; CD, CQ).</math>
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Therefore,
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<cmath>\frac{(PX)(MQ)}{(PQ)(MX)} = \frac{(PM)(YQ)}{(PQ)(YM)}.</cmath>
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Since <math>MQ = PM</math>,
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<cmath>\frac{MX}{YM} = \frac{XP}{QY}.</cmath>
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Moreover,
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<cmath>\frac{MX + PX}{YM + QY} = 1,</cmath>
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so <math>MX = YM,</math> as desired.
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<math>\blacksquare</math>.
  
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==Related Reading==
 
http://agutie.homestead.com/FiLEs/GeometryButterfly.html
 
http://agutie.homestead.com/FiLEs/GeometryButterfly.html
  
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http://www.mathematik.uni-muenchen.de/~fritsch/butterfly.pdf
  
 
==See also==
 
==See also==
 
[[Category:Geometry]]
 
[[Category:Geometry]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]
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[[Midpoint]]

Latest revision as of 11:32, 26 January 2015

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

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