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 A B C {\displaystyle ABC} points D {\displaystyle D} , E {\displaystyle E} , and F {\displaystyle F} lie on segments B C {\displaystyle BC} , C A {\displaystyle CA} , and A B {\displaystyle AB} , then writing C D B D = x {\displaystyle {\tfrac {CD}{BD}}=x} , A E C E = y {\displaystyle {\tfrac {AE}{CE}}=y} , and B F A F = z {\displaystyle {\tfrac {BF}{AF}}=z} , the signed area of the triangle formed by the cevians A D {\displaystyle AD} , B E {\displaystyle BE} , and C F {\displaystyle CF} is
S A B C ⋅ ( x y z − 1 ) 2 ( x y + y + 1 ) ( y z + z + 1 ) ( z x + x + 1 ) , {\displaystyle S_{ABC}\cdot {\frac {(xyz-1)^{2}}{(xy+y+1)(yz+z+1)(zx+x+1)}},}
where S A B C {\displaystyle S_{ABC}} is the area of the triangle A B C {\displaystyle 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 {\displaystyle x=y=z=2} has become popularized as the one-seventh area triangle. The x = y = z = 1 {\displaystyle x=y=z=1} case implies that the three medians are concurrent (through the centroid).
Proof
Suppose that the area of triangle A B C {\displaystyle ABC} is 1. For triangle A B D {\displaystyle ABD} and line F R C {\displaystyle FRC} , Menelaus's theorem implies
A F F B × B C C D × D R R A = − 1 {\displaystyle {\frac {AF}{FB}}\times {\frac {BC}{CD}}\times {\frac {DR}{RA}}=-1} . Then D R R A = B F F A × C D C B = z x x + 1 {\displaystyle {\frac {DR}{RA}}={\frac {BF}{FA}}\times {\frac {CD}{CB}}={\frac {zx}{x+1}}} . Thus the area of triangle A R C {\displaystyle ARC} is
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