# 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 states that if in triangle ABC points D, E, and F lie on segments BC, CA, and AB, then writing \(\dfrac{CD}{BD} = x\), \(\dfrac{AE}{CE} = y\), and\( \dfrac{BF}{AF} = z\), the signed area of the triangle ie \(\Delta PQR\) formed by the cevians AD, BE, and CF is the area of \(\Delta ABC\) times \(\dfrac{(xyz - 1)^2}{(xy + y + 1)(yz + z + 1)(zx + x + 1)}\).

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,

\(\dfrac{FA}{BF}\times\dfrac{CB}{DC}\times\dfrac{DR}{RA}=1\\\Rightarrow\dfrac{DR}{RA}=\dfrac{BF}{FA}\times\dfrac{DC}{CB}=\dfrac{BF}{FA}\times\dfrac{DC}{BD+DC}=\dfrac{cz}{c}\times\dfrac{ax}{a+ax}=\dfrac{zx}{1+x}\)

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 \(\Delta CQB\)

\(S'=S-\dfrac{xS}{zx+x+1}-\dfrac{yS}{xy+y+1}-\dfrac{zS}{yz+z+1}\\=S\left (1-\left (\dfrac{xS}{zx+x+1}+\dfrac{yS}{xy+y+1}+\dfrac{zS}{yz+z+1}\right )\right )\\=\dfrac{(xyz - 1)^2S}{(xy + y + 1)(yz + z + 1)(zx + x + 1)}\)

**Note:** \(x=y=z=1\) is a special case of medians which are concurrent, thus the area is \(0\)

## Let the area of an \(\Delta ABC\) be \(35\) and \(D,E,F\) be the points on \(BC,AC,AB\) 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, \(x=y=z=2\)

Substituting \(x,y,z\),

\(\Delta PQR=35\times\dfrac{7^2}{(7\times 7\times 7)}=5\)