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  "body": "\n## Statement\n\nLet $ABC$ be a [[triangle|0CTNFGYMM2]] in the Euclidean plane with area $S$.\nLet $D$, $E$, $F$ be points in the interiors of the sides $BC$, $CA$, $AB$, and let $x = BD/DC$, $y = CE/EA$ and $z = AF/FB$.\nThen the three segments $AD$, $BE$, $CF$ bound a triangle whose [[area|065JWMZW86]] is\n$$S \\cdot \\frac{(xyz - 1)^2}{(xy + x + 1)(yz + y + 1)(zx + z + 1)}.$$\nWhen $xyz = 1$, this triangle degenerates to a point.\n\n## Remarks\n\nA proof places the triangle in barycentric coordinates, computes the three intersection points of the cevians, and takes the ratio of determinants.\nExample: with $x = y = z = 2$, the area is $S \\cdot 49/343 = S/7$, the well-known one-seventh triangle.\nThe case $xyz = 1$ is exactly the condition of [[Ceva's theorem|0KN93VDSXA]] for the three cevians to be concurrent, and Routh's formula gives area $0$ there.\n\n## Explanations\n\n### L1 human\n\nDraw three lines from the corners of a triangle to points on the opposite sides.\nIn general the three lines enclose a small triangle in the middle.\nRouth's theorem tells exactly how big that small triangle is, using only the ratios in which the sides are cut.\n\n### L2 human\n\nRouth's theorem refines Ceva's theorem: Ceva says when the inner triangle is a point, and Routh measures it in all other cases.\nBecause affine maps preserve ratios of areas and ratios along lines, it is enough to prove it for one triangle, for example $A = (0, 0)$, $B = (1, 0)$, $C = (0, 1)$.\nThe theorem is a good exercise in barycentric coordinates and in the use of the [[determinant|00KDZRW4C0]] to compute areas.\n",
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