# 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 \(\Delta CQB):\)

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

**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 \(\Delta PQR\).

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

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