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Difference between revisions of "Routh's Theorem"

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In [[triangle]] <math>ABC</math>, <math>D</math>, <math>E</math> and <math>F</math> are points on sides <math>BC</math>, <math>AC</math>, and <math>AB</math>, respectively. Let <math>r=\frac{AF}{AB}</math>, <math>s=\frac{BD}{BC}</math>, and <math>=\frac{CE}{CA}</math>. Let <math>G</math> be the intersection of <math>AD</math> and <math>BC</math>, <math>H</math> be the intersection of <math>BE</math> and <math>CF</math>, and <math>I</math> be the intersection of <math>CF</math> and <math>AD</math>. Then, '''Routh's Theorem''' states that  
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In [[triangle]] <math>ABC</math>, <math>D</math>, <math>E</math> and <math>F</math> are points on sides <math>BC</math>, <math>AC</math>, and <math>AB</math>, respectively. Let <math>r=\frac{AF}{FB}</math>, <math>s=\frac{BD}{DC}</math>, and <math>t=\frac{CE}{AE}</math>. Let <math>G</math> be the intersection of <math>AD</math> and <math>BC</math>, <math>H</math> be the intersection of <math>BE</math> and <math>CF</math>, and <math>I</math> be the intersection of <math>CF</math> and <math>AD</math>. Then, '''Routh's Theorem''' states that  
  
 
<cmath>[GHI]=\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]</cmath>
 
<cmath>[GHI]=\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]</cmath>
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==Proof==
 
==Proof==
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{{incomplete|proof}}
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Scale [[triangle]]<math>ABC</math>'s area down so that it becomes 1. We can then use Menelaus's Theorem on [[triangle]] <math>ABD</math> and line <math>FHC</math>.
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<math>\frac{AF}{FB}\times\frac{BC}{CD}\times\frac{DG}{GA}= 1</math>
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This means <math>\frac{DG}{GA}= \frac{BF}{FA}\times\frac{DC}{CB} = \frac{rs}{s+1}</math>
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Full proof: https://en.wikipedia.org/wiki/Routh%27s_theorem
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== See also ==
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* [[Menelaus' Theorem]]
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*[[Ceva's Theorem]]
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{{stub}}
 
 
[[Category:Geometry]]
 
[[Category:Geometry]]
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[[Category:Theorems]]
 
[[Category:Theorems]]
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[[Category:Geometry]]
 
[[Category:Definition]]
 
[[Category:Definition]]
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{{stub}}

Latest revision as of 23:02, 23 March 2021

In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\frac{AF}{FB}$, $s=\frac{BD}{DC}$, and $t=\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that

\[[GHI]=\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]\]

[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW); label("$G$",G,N); label("$H$",H,N); label("$I$",I,SW);[/asy]

Proof

Scale triangle$ABC$'s area down so that it becomes 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\frac{AF}{FB}\times\frac{BC}{CD}\times\frac{DG}{GA}= 1$ This means $\frac{DG}{GA}= \frac{BF}{FA}\times\frac{DC}{CB} = \frac{rs}{s+1}$

Full proof: https://en.wikipedia.org/wiki/Routh%27s_theorem

See also

This article is a stub. Help us out by expanding it.

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