Routh's Theorem

Routh's theorem determines the ratio of areas between a given triangle and a triangle formed by the pairwise intersections of three cevians.

The theorem goes as follows:

In triangle $ABC,$ if points $D, E,$ and $F$ lie on segments $BC, CA,$ and $AB,$ respectively, then writing $\frac{CD}{BD} = x$, $\frac{AE}{CE} = y$, and$\frac{BF}{AF} = z$, the (red) area of $\triangle PQR$ formed by cevians $AD, BE,$ and $CF$ is equal to the area of $\triangle ABC$ times

$$\frac{(xyz - 1)^2}{(xy + y + 1)(yz + z + 1)(zx + x + 1)}.$$

Let the area of $\Delta ABC$ be $S$ and that of $\Delta PQR$ be $S'.$

Applying Menelaus's theorem on $\Delta ABD$ and line $FRC$, we get

$$\dfrac{FA}{BF}\times\dfrac{CB}{DC}\times\dfrac{DR}{RA}=1\implies \dfrac{DR}{RA}=\dfrac{BF}{FA}\times\dfrac{DC}{CB}=\dfrac{BF}{FA}\times\dfrac{DC}{BD+DC}=\dfrac{cz}{c}\times\dfrac{ax}{a+ax}=\dfrac{zx}{1+x}.$$

$($Area of $\Delta ADC)=\dfrac{DC}{BC}\times$$($Area of $\Delta ABC)=\dfrac{xS}{1+x}.$
$($Area of $\Delta ARC)=\dfrac{AR}{AD}\times$$($Area of $\Delta ADC)=\dfrac{xS}{zx+x+1}.$
By similar arguments, $($Area of $\Delta BPA)=\dfrac{yS}{xy+y+1}$ and $($Area of $\Delta CQB)=\dfrac{zS}{yz+z+1}.$

Now, $($Area of $\Delta PQR)=$$($Area of $\Delta ABC)-$$($Area of $\Delta ARC)-$$($Area of $\Delta BPA)-$$($Area of $\Delta CQB):$

$$\begin{aligned} S' &=S-\dfrac{xS}{zx+x+1}-\dfrac{yS}{xy+y+1}-\dfrac{zS}{yz+z+1}\\ &=S\left [1-\left (\dfrac{xS}{zx+x+1}+\dfrac{yS}{xy+y+1}+\dfrac{zS}{yz+z+1}\right )\right]\\ &=\dfrac{(xyz - 1)^2S}{(xy + y + 1)(yz + z + 1)(zx + x + 1)}.\ _\square \end{aligned}$$

Note: $x=y=z=1$ is a special case of medians which are concurrent, and thus the area is $0.$

Let the area of $\Delta ABC$ be $35$ and $D,E,$ and $F$ be points on $BC,AC,$ and $AB,$ respectively, such that $CD=2BD,AE=2CE, BF=2AF$.

Calculate the area of $\Delta PQR$.

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By using Routh's theorem, $\Delta PQR=\Delta ABC\times\dfrac{(xyz - 1)^2}{(xy + y + 1)(yz + z + 1)(zx + x + 1)}.$

In this case, $x=y=z=2.$ So, substituting $x,y,z$ gives

$$\Delta PQR=35\times\dfrac{7^2}{7\times 7\times 7}=5.\ _\square$$