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  "body": "\n## Proof\n\nWe identify the plane with the [[complex numbers|09FTAET5Z4]] and use an algebraic identity.\n\n1. Let $S$ be the common point of the diagonals, with $S = (1 - s)A + sC = (1 - t)B + tD$ and $0 \u003c s, t \u003c 1$. So $A \\neq C$, $B \\neq D$, and $S$ is none of the four points.\n   The points are [[concyclic|049DCJZ448]], so they lie on a [[circle|060W5CBZP6]] with a centre $O$ and a radius $r$. Since $A \\neq C$, we have $r \u003e 0$.\n   A line $\\set{Y + \\tau v}$ with $v \\neq 0$ meets the circle where $\\norm{Y - O + \\tau v}^2 = r^2$. This is a quadratic equation in $\\tau$ with leading coefficient $\\norm{v}^2 \u003e 0$, so there are at most two such points.\n   Suppose that $B$ or $D$ is equal to $A$ or to $C$. Then the lines $AC$ and $BD$ share $S$ and a vertex, so they are equal. This line contains $A$, $B$, $C$, $D$, so $\\set{B, D} = \\set{A, C}$.\n   If $B = A$ and $D = C$, or if $B = C$ and $D = A$, then both sides of the claim equal $d(A, C)^2$. From now on, the four points are distinct.\n2. Put $a = A - O$, $b = B - O$, $c = C - O$, $e = D - O$. Then $d(X, Y) = \\abs{x - y}$ for the [[Euclidean distance|0NPPGHQNXQ]], with the [[modulus|0KW2X0TFBX]].\n3. For all complex numbers, $(a - c)(b - e) = (a - b)(c - e) + (a - e)(b - c)$. Indeed, both sides expand to $ab - ae - bc + ce$.\n4. For real $x$, $y$, Euler's formula gives $e^{ix} - e^{iy} = 2i\\sin\\frac{x - y}{2}\\, e^{i(x + y)/2}$. This follows from $e^{ih} - e^{-ih} = 2i\\sin h$ by [[Euler's formula|04R3CY41VJ]].\n5. By the [[polar form|0BRXFNQ5JV]], write $a = re^{i\\alpha}$ and $c = re^{i\\gamma}$. Since $C \\neq A$, we may choose $\\gamma$ with $\\alpha \u003c \\gamma \u003c \\alpha + 2\\pi$.\n6. Let $X$ be a point of the circle other than $A$. Write $X - O = re^{i\\theta}$ with $\\alpha \u003c \\theta \u003c \\alpha + 2\\pi$. Put $g(X) = \\det(C - A, X - A)$, with the [[determinant|00KDZRW4C0]].\n7. For $u = u_1 + iu_2$ and $w = w_1 + iw_2$, the imaginary part of $\\bar u\\, w$ is $u_1 w_2 - u_2 w_1 = \\det(u, w)$. So $g(X)$ is the imaginary part of $\\overline{(c - a)}\\,(x - a)$, with the [[complex conjugate|0FTH2E9TAS]] and $x = X - O$.\n8. By steps 4 and 6 and the [[product of exponentials|0WM7DJCM6G]], $\\overline{(c - a)}\\,(x - a) = 4r^2 \\sin\\frac{\\gamma - \\alpha}{2}\\sin\\frac{\\theta - \\alpha}{2}\\, e^{i(\\theta - \\gamma)/2}$.\n9. So $g(X) = 4r^2 \\sin\\frac{\\gamma - \\alpha}{2}\\sin\\frac{\\theta - \\alpha}{2}\\sin\\frac{\\theta - \\gamma}{2}$. The first two sines are positive, since their arguments lie in $(0, \\pi)$.\n10. The third sine is negative for $\\theta \u003c \\gamma$ and positive for $\\theta \u003e \\gamma$, since its argument lies in $(-\\pi, \\pi)$. So $g(X) \u003c 0$ for $\\theta \\in (\\alpha, \\gamma)$ and $g(X) \u003e 0$ for $\\theta \\in (\\gamma, \\alpha + 2\\pi)$.\n11. By step 1, $S = (1 - t)B + tD$ with $0 \u003c t \u003c 1$, and $S$ is on the [[line|0MP9BCF6DP]] $AC$. So $g(S) = 0$. The determinant is linear in its second column, so $(1 - t)g(B) + t g(D) = 0$.\n12. By step 10, $g(B) \\neq 0$ and $g(D) \\neq 0$, since $B$ and $D$ differ from $A$ and $C$. By step 11, $g(B)$ and $g(D)$ have opposite signs.\n13. The claimed equation does not change if we swap the names $B$ and $D$. So by step 10 we may assume $b = re^{i\\beta}$ and $e = re^{i\\delta}$ with $\\alpha \u003c \\beta \u003c \\gamma \u003c \\delta \u003c \\alpha + 2\\pi$.\n14. Put $\\sigma = \\frac{\\alpha + \\beta + \\gamma + \\delta}{2}$. By step 4, $(a - b)(c - e) = -4r^2 e^{i\\sigma} s_1$, with $s_1 = \\sin\\frac{\\alpha - \\beta}{2}\\sin\\frac{\\gamma - \\delta}{2}$.\n15. In the same way, $(a - e)(b - c) = -4r^2 e^{i\\sigma} s_2$ with $s_2 = \\sin\\frac{\\alpha - \\delta}{2}\\sin\\frac{\\beta - \\gamma}{2}$.\n16. By step 13, the numbers $\\frac{\\alpha - \\beta}{2}$, $\\frac{\\gamma - \\delta}{2}$, $\\frac{\\alpha - \\delta}{2}$ and $\\frac{\\beta - \\gamma}{2}$ all lie in $(-\\pi, 0)$. So each of these sines is negative, and $s_1 \u003e 0$, $s_2 \u003e 0$.\n17. By steps 3, 14 and 15, $(a - c)(b - e) = -4r^2 e^{i\\sigma}(s_1 + s_2)$. Taking moduli and using $\\abs{e^{i\\sigma}} = 1$, we get $\\abs{a - c}\\abs{b - e} = 4r^2(s_1 + s_2)$.\n18. Also by steps 14 and 15, $\\abs{a - b}\\abs{c - e} = 4r^2 s_1$ and $\\abs{a - e}\\abs{b - c} = 4r^2 s_2$.\n19. By steps 2, 17 and 18, $d(A, C)\\, d(B, D) = d(A, B)\\, d(C, D) + d(B, C)\\, d(A, D)$.\n\n## Idea\n\nThe identity of step 3 holds for any four complex numbers, so the triangle inequality gives Ptolemy's inequality. For points in convex order on a circle, the two products on the right point in the same direction. Then the moduli add exactly, and the inequality becomes an equation.\n",
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