Difference between revisions of "Routh's Theorem"

m
Line 1: Line 1:
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>t=\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  
+
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{BF}{AF}</math>, <math>s=\frac{CD}{BD}</math>, and <math>t=\frac{AE}{CE}</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>

Revision as of 23:34, 23 May 2013

In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\frac{BF}{AF}$, $s=\frac{CD}{BD}$, and $t=\frac{AE}{CE}$. 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

Template:Incomplete

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

See also