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  "body": "\n## Proof\n\nWe measure signed distances to $\\ell$ with a determinant.\n\n1. Write the [[line|0MP9BCF6DP]] $\\ell$ as $\\set{p + tv : t \\in \\mathbb{R}}$, with a point $p$ and a nonzero vector $v$. For $X \\in \\mathbb{R}^2$ put $f(X) = \\det(v, X - p)$, the [[determinant|00KDZRW4C0]] of the $2 \\times 2$ matrix with the columns $v$ and $X - p$.\n2. For a vector $u$, $\\det(v, u) = 0$ if and only if $u$ is a multiple of $v$, because $v \\neq 0$. So $f(X) = 0$ if and only if $X \\in \\ell$.\n3. The determinant is linear in its second column. So for an [[affine combination|0KJH6SRFZK]] $X = (1 - s)Y + sZ$ we have $f(X) = (1 - s) f(Y) + s f(Z)$.\n4. Put $a = f(A)$, $b = f(B)$ and $c = f(C)$. They are nonzero by step 2, because $\\ell$ does not pass through the vertices $A$, $B$, $C$ of the [[triangle|0CTNFGYMM2]].\n5. Let $X = (1 - s)Y + sZ$ with $Y \\neq Z$, $s \\neq 1$. Then $X - Y = s(Z - Y)$ and $Z - X = (1 - s)(Z - Y)$. So the signed ratio is $\\frac{YX}{XZ} = \\frac{s}{1 - s}$.\n\n**The direct statement.**\n\n6. The point $D$ is on the line $BC$, so $D = (1 - s)B + sC$ for some $s \\in \\mathbb{R}$. Since $D \\in \\ell$, steps 2 and 3 give $(1 - s)b + sc = 0$.\n7. If $s = 1$, this gives $c = 0$, against step 4. So $s \\neq 1$, and also $s \\neq 0$ since $b \\neq 0$.\n8. By steps 5 and 6, $\\frac{BD}{DC} = \\frac{s}{1 - s} = -\\frac{b}{c}$.\n9. In the same way, $\\frac{CE}{EA} = -\\frac{c}{a}$ and $\\frac{AF}{FB} = -\\frac{a}{b}$.\n10. Multiplying, $\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA} = \\left(-\\frac{a}{b}\\right)\\left(-\\frac{b}{c}\\right)\\left(-\\frac{c}{a}\\right) = -1$.\n\n**The converse.**\n\n11. Let $D$, $E$, $F$ be as in the converse. Put $x = \\frac{BD}{DC}$, $y = \\frac{CE}{EA}$ and $z = \\frac{AF}{FB}$, so $xyz = -1$.\n12. 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 $D \\neq B$.\n13. In the same way, $(1 + y)E = C + yA$ and $(1 + z)F = A + zB$.\n14. Then $(1 + y)E - y(1 + z)F = C + yA - yA - yzB = C - yzB$.\n15. Since $xyz = -1$ and $x \\neq 0$, we have $-yz = \\frac{1}{x}$. So the right side of step 14 is $C + \\frac{1}{x}B = \\frac{1 + x}{x} D$ by step 12.\n16. The coefficients on the left of step 14 add up to $1 + y - y - yz = 1 - yz = \\frac{1 + x}{x}$.\n17. Divide step 14 by $\\frac{1 + x}{x} \\neq 0$. This gives $D = \\lambda E + \\mu F$ with $\\lambda + \\mu = 1$, so $D$ is an affine combination of $E$ and $F$.\n18. The points $E$ and $F$ are distinct, because $E = F$ would lie on $CA$ and $AB$, so it would be the vertex $A$. So $E$ and $F$ span a line, and $D = F + \\lambda(E - F)$ lies on it.\n19. So $D$, $E$ and $F$ are [[collinear|0AFNNTRTJE]].\n\n## Idea\n\nThe function $f$ is a signed distance to $\\ell$, up to a constant factor. A point of a side lies on $\\ell$ exactly when the signed distances of the two endpoints balance. So each ratio is minus a quotient of two of them. The three quotients cancel and leave the three minus signs.\n",
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