Routh's theorem



In geometry, 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 $$\tfrac{CD}{BD} = x$$, $$\tfrac{AE}{CE} = y$$, and $$\tfrac{BF}{AF} = z$$, the signed area of the triangle formed by the cevians $$AD$$, $$BE$$, and $$CF$$ is
 * $$S_{ABC} \frac{(xyz - 1)^2}{(xy + y + 1)(yz + z + 1)(zx + x + 1)},$$

where $$S_{ABC}$$ is the area of the triangle $$ABC$$.

This theorem was given by Edward John Routh on page 82 of his Treatise on Analytical Statics with Numerous Examples in 1896. The particular case $$ x = y = z = 2$$ has become popularized as the one-seventh area triangle. The $$ x = y = z = 1$$ case implies that the three medians are concurrent (through the centroid).

Proof


Suppose that the area of triangle $$ABC$$ is 1. For triangle $$ABD$$ and line $$FRC$$ using Menelaus's theorem, We could obtain:
 * $$\frac{AF}{FB} \times \frac{BC}{CD} \times \frac{DR}{RA} = 1$$

Then $$\frac{DR}{RA} = \frac{BF}{FA} \times \frac{DC}{CB} = \frac{zx}{x+1}$$ So the area of triangle $$ARC$$ is:
 * $$S_{ARC} = \frac{AR}{AD} S_{ADC} = \frac{AR}{AD} \times \frac{DC}{BC} S_{ABC} = \frac{x}{zx+x+1}$$

Similarly, we could know: $$S_{BPA} = \frac{y}{xy+y+1}$$ and $$S_{CQB} = \frac{z}{yz+z+1}$$ Thus the area of triangle $$PQR$$ is:
 * $$\begin{align}

S_{PQR} &= S_{ABC} - S_{ARC} - S_{BPA} - S_{CQB} \\ &= 1 - \frac{x}{zx+x+1} - \frac{y}{xy+y+1} - \frac{z}{yz+z+1} \\ &=\frac{(xyz - 1)^2}{(xz + x + 1)(yx + y + 1)(zy + z + 1)}. \end{align}$$