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  "body": "\n## Proof\n\nWe use barycentric coordinates with respect to $A$, $B$, $C$.\n\n1. The points $A$, $B$, $C$ of the [[triangle|0CTNFGYMM2]] are not [[collinear|0AFNNTRTJE]]. So the vectors $B - A$ and $C - A$ are [[linearly independent|0CRMGPAH83]].\n2. Two linearly independent vectors of $\\mathbb{R}^2$ form a [[basis|096V7H13H9]]. So each point $X$ has a unique form $X = A + u(B - A) + v(C - A)$.\n3. Equivalently, each point $X$ is a unique [[affine combination|0KJH6SRFZK]] $X = \\alpha A + \\beta B + \\gamma C$ with $\\alpha + \\beta + \\gamma = 1$. We call $(\\alpha, \\beta, \\gamma)$ the coordinates of $X$.\n4. The [[line|0MP9BCF6DP]] $BC$ is the set of the points $(1 - s)B + sC$ with $s \\in \\mathbb{R}$. By the uniqueness in step 3, a point lies on $BC$ if and only if $\\alpha = 0$. In the same way, $CA$ is $\\beta = 0$ and $AB$ is $\\gamma = 0$.\n5. Let $X = \\alpha A + \\beta B + \\gamma C$ be a point of $AB$, so $\\gamma = 0$ and $\\alpha + \\beta = 1$, and let $X \\neq B$, so $\\alpha \\neq 0$. Then $X - A = \\beta(B - A)$ and $B - X = \\alpha(B - A)$. Hence the signed ratio is $\\frac{AX}{XB} = \\frac{\\beta}{\\alpha}$.\n\n**The direct statement.**\n\n6. Let $O = \\alpha A + \\beta B + \\gamma C$ with $\\alpha + \\beta + \\gamma = 1$. Since $O$ is not on $BC$, $CA$, $AB$, step 4 gives $\\alpha \\neq 0$, $\\beta \\neq 0$ and $\\gamma \\neq 0$.\n7. A point of the line $AO$ has the form $(1 - t)A + tO = (1 - t + t\\alpha)A + t\\beta B + t\\gamma C$. Its first coordinate is $1 - t(\\beta + \\gamma)$.\n8. If $\\beta + \\gamma = 0$, then this coordinate is $1$ for all $t$. Then by step 4 the line $AO$ does not meet $BC$, against the hypothesis. So $\\beta + \\gamma \\neq 0$.\n9. By step 4, the point $D$ is the point of step 7 with $t = \\frac{1}{\\beta + \\gamma}$. Thus $D = \\frac{\\beta}{\\beta + \\gamma} B + \\frac{\\gamma}{\\beta + \\gamma} C$, and $D \\neq C$ since $\\beta \\neq 0$.\n10. By step 5 applied to the side $BC$, we get $\\frac{BD}{DC} = \\frac{\\gamma}{\\beta}$.\n11. In the same way, $\\alpha + \\beta \\neq 0$ and $\\gamma + \\alpha \\neq 0$. Also $F = \\frac{\\alpha A + \\beta B}{\\alpha + \\beta}$ and $E = \\frac{\\gamma C + \\alpha A}{\\gamma + \\alpha}$. So $\\frac{AF}{FB} = \\frac{\\beta}{\\alpha}$ and $\\frac{CE}{EA} = \\frac{\\alpha}{\\gamma}$.\n12. Multiplying, $\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA} = \\frac{\\beta}{\\alpha} \\cdot \\frac{\\gamma}{\\beta} \\cdot \\frac{\\alpha}{\\gamma} = 1$.\n\n**The converse.**\n\n13. Let $D$, $E$, $F$ be as in the converse. Put $x = \\frac{BD}{DC}$, $y = \\frac{CE}{EA}$ and $z = \\frac{AF}{FB}$, so $xyz = 1$.\n14. From $D - B = x(C - D)$ we get $(1 + x)D = B + xC$. If $x = -1$, this gives $B = C$, which is false. So $1 + x \\neq 0$. Also $x \\neq 0$, because $D \\neq B$.\n15. In the same way, $(1 + y)E = C + yA$ and $(1 + z)F = A + zB$, with $y, z \\notin \\set{0, -1}$.\n16. Put $S = 1 + z + zx$ and $w = A + zB + zxC$. Using step 14, $z(1 + x)D = zB + zxC$. Hence $w = A + z(1 + x)D$.\n17. Using $zxy = 1$ and step 15, $zx(1 + y)E = zxC + A$. Hence $w = zB + zx(1 + y)E$. Note $zx(1 + y) = zx + 1$.\n18. Using step 15, $w = zxC + (1 + z)F$.\n19. Case $S \\neq 0$. Put $Q = w / S$. By step 16, $Q = \\frac{1}{S}A + \\frac{z(1 + x)}{S}D$, an affine combination of $A$ and $D$ since the weights add up to $1$. So $Q$ lies on the line $AD$.\n20. By steps 17 and 18 in the same way, $Q$ lies on the lines $BE$ and $CF$. So the three lines meet in the point $Q$.\n21. Case $S = 0$. Then $w = (A + zB + zxC) - S A = z(B - A) + zx(C - A)$. This is a nonzero vector by step 1, since $z \\neq 0$ by step 15.\n22. Since $S = 0$, step 16 gives $z(1 + x)(D - A) = w - SA = w$. So $w$ is a direction vector of the line $AD$, as $z(1 + x) \\neq 0$.\n23. In the same way, steps 17 and 18 give $(1 + zx)(E - B) = w$ and $(1 + z)(F - C) = w$. Here $1 + zx = zx(1 + y) \\neq 0$ and $1 + z \\neq 0$.\n24. So the lines $AD$, $BE$ and $CF$ all have the direction $w$, and they are parallel.\n\n## Idea\n\nBarycentric coordinates turn each ratio on a side into a quotient of two coordinates of the point $O$. The product of the three quotients telescopes to $1$. For the converse, the weights $1$, $z$, $zx$ build the common point directly. When these weights add up to $0$, the same computation gives a common direction.\n",
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