# Routh's Theorem

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

Assume triangle$ABC$'s area to be 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 $\frac{DG}{GA}= \frac{BF}{FA}\times\frac{DC}{CB} = \frac{rs}{s+1}$