Difference between revisions of "Routh's Theorem"
(New page: 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{A...) |
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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}{ | + | 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> | ||
| + | |||
| + | <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== | ==Proof== | ||
| − | {{ | + | 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>. |
| + | <math>\frac{AF}{FB}\times\frac{BC}{CD}\times\frac{DG}{GA}= 1</math> | ||
| + | This means <math>\frac{DG}{GA}= \frac{BF}{FA}\times\frac{DC}{CB} = \frac{rs}{s+1}</math> | ||
| + | |||
| + | Full proof: https://en.wikipedia.org/wiki/Routh%27s_theorem | ||
| + | |||
| + | == See also == | ||
| + | * [[Menelaus' Theorem]] | ||
| + | *[[Ceva's Theorem]] | ||
| + | |||
| − | |||
[[Category:Geometry]] | [[Category:Geometry]] | ||
| + | |||
[[Category:Theorems]] | [[Category:Theorems]] | ||
| + | [[Category:Geometry]] | ||
[[Category:Definition]] | [[Category:Definition]] | ||
| + | |||
| + | {{stub}} | ||
Latest revision as of 23:02, 23 March 2021
In triangle 
, 
, 
and 
are points on sides 
, 
, and 
, respectively. Let 
, 
, and 
. Let 
be the intersection of 
and , 
be the intersection of 
and 
, and 
be the intersection of and . Then, Routh's Theorem states that
![\[[GHI]=\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]\]](http://latex.artofproblemsolving.com/1/a/a/1aa432e9368f5d0d8d3af8c51676097923f806c0.png)
![[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]](http://latex.artofproblemsolving.com/2/3/9/239cda04d81c236a617bf26c6c34e361b13d4f18.png)
Proof
Scale triangle's area down so that it becomes 1. We can then use Menelaus's Theorem on triangle 
and line 
.
This means 
Full proof: https://en.wikipedia.org/wiki/Routh%27s_theorem
See also
This article is a stub. Help us out by expanding it.
