Routh's Theorem

Revision as of 15:16, 14 March 2017 by Halt catchfire (talk | contribs) (Proof)

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


[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]


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 $\frac{DG}{GA}= \frac{BF}{FA}\times\frac{DC}{CB} = \frac{rs}{s+1}$

Full proof:

See also

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