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  "body": "\n## Proof\n\nWe write all points with vectors from $S$ and compare coefficients.\n\n1. Two distinct [[lines|0MP9BCF6DP]] meet in at most one point, because exactly one line passes through two distinct points. So $\\ell_1 \\cap \\ell_2 = \\set{S}$.\n2. Write $\\ell_1 = \\set{P_0 + \\tau w : \\tau \\in \\R}$ with $w \\neq 0$. Then $S = P_0 + \\tau_1 w$ and $A = P_0 + \\tau_2 w$ with $\\tau_1 \\neq \\tau_2$, since $A \\neq S$.\n3. Let $p = A - S = (\\tau_2 - \\tau_1) w \\neq 0$. Each point $P_0 + \\tau w$ equals $S + \\frac{\\tau - \\tau_1}{\\tau_2 - \\tau_1} p$. So $\\ell_1 = \\set{S + \\sigma p : \\sigma \\in \\R}$.\n4. In the same way, with $q = C - S \\neq 0$, we get $\\ell_2 = \\set{S + \\sigma q : \\sigma \\in \\R}$.\n5. The vectors $p$ and $q$ are [[linearly independent|0CRMGPAH83]]. Otherwise $q = \\mu p$ for a real $\\mu$, so $C \\in \\ell_1$. Then $C \\in \\ell_1 \\cap \\ell_2$ and $C \\neq S$, against step 1.\n6. By steps 3 and 4, $B = S + s p$ and $D = S + t q$ for real numbers $s$ and $t$. Since $B \\neq S$ and $D \\neq S$, we have $s \\neq 0$ and $t \\neq 0$.\n7. By step 5, $C - A = q - p \\neq 0$ and $D - B = t q - s p \\neq 0$. So $C - A$ and $D - B$ are direction vectors of the lines $AC$ and $BD$.\n8. Two lines are parallel when they have the same direction vectors, up to a nonzero factor. So $D - B = \\mu (C - A)$ for a real $\\mu$.\n9. Step 8 says $t q - s p = \\mu q - \\mu p$, that is $(t - \\mu) q + (\\mu - s) p = 0$. By step 5, $t = \\mu$ and $s = \\mu$. So $s = t$.\n10. With the [[Euclidean distance|0NPPGHQNXQ]] and the [[length|0XKBKGTCPC]], $\\abs{SB} = \\norm{s p} = \\abs{s} \\, \\abs{SA}$ and $\\abs{SD} = \\norm{s q} = \\abs{s} \\, \\abs{SC}$.\n11. Also $\\abs{BD} = \\norm{s q - s p} = \\abs{s} \\, \\norm{q - p} = \\abs{s} \\, \\abs{AC}$.\n12. All of $\\abs{SB}$, $\\abs{SD}$ and $\\abs{BD}$ are positive, by steps 6 and 7. Dividing, each of the three quotients $\\frac{\\abs{SA}}{\\abs{SB}}$, $\\frac{\\abs{SC}}{\\abs{SD}}$ and $\\frac{\\abs{AC}}{\\abs{BD}}$ equals $\\frac{1}{\\abs{s}}$.\n13. Hence $\\dfrac{\\abs{SA}}{\\abs{SB}} = \\dfrac{\\abs{SC}}{\\abs{SD}} = \\dfrac{\\abs{AC}}{\\abs{BD}}$.\n\n## Idea\n\nFrom $S$, the points $B$ and $D$ are multiples $s p$ and $t q$ of the vectors to $A$ and $C$. The parallel condition forces $s = t$, because $p$ and $q$ point in independent directions. So one scaling by $s$ maps $A$, $C$ to $B$, $D$, and every length scales by $\\abs{s}$.\n",
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