Difference between revisions of "Routh's Theorem"
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==Proof== | ==Proof== | ||
| − | <math> | + | Assume [[triangle]]<math>ABC</math>'s area to be 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 $\frac{DG}{GA} = \frac{BR}{FA}\times\frac{DC}{CB} = \frac{rs}{s+1} | ||
| + | |||
== See also == | == See also == | ||
* [[Menelaus' Theorem]] | * [[Menelaus' Theorem]] | ||
Revision as of 17:10, 18 August 2014
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
Assume triangle's area to be 1. We can then use Menelaus's Theorem on triangle
and line 
.
This means $\frac{DG}{GA} = \frac{BR}{FA}\times\frac{DC}{CB} = \frac{rs}{s+1}
See also
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