Difference between revisions of "Routh's Theorem"

Latest revision as of 23:02, 23 March 2021

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

Scale triangle $ABC$ 's area down so that it becomes 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: https://en.wikipedia.org/wiki/Routh%27s_theorem

@@ Line 32: / Line 32: @@
 ==Proof==
-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>.
+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>

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Difference between revisions of "Routh's Theorem"

Latest revision as of 23:02, 23 March 2021

Proof

See also