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  "body": "\n## Proof\n\nWe expand a squared length, which proves both directions at once.\n\n1. Let $x = A - B$ and $y = C - B$. With the [[Euclidean distance|0NPPGHQNXQ]] and the [[length|0XKBKGTCPC]], $d(A, B) = \\norm{x}$, $d(B, C) = \\norm{y}$ and $d(A, C) = \\norm{x - y}$.\n2. The [[dot product|0S52TNNTH9]] is bilinear and symmetric. So $\\norm{x - y}^2 = \\norm{x}^2 + \\norm{y}^2 - 2 \\, x \\cdot y$.\n3. By steps 1 and 2, $d(A, C)^2 = d(A, B)^2 + d(B, C)^2$ holds if and only if $x \\cdot y = 0$.\n4. The vertices of the [[triangle|0CTNFGYMM2]] are not collinear, so they are distinct. Thus $x \\neq 0$ and $y \\neq 0$.\n5. By definition, the [[angle|0GJAJC6TNZ]] $\\angle ABC$ is the angle between $x$ and $y$. Since both are nonzero, $x \\cdot y = 0$ holds [[if and only if this angle is $\\pi / 2$|03NK6RSRQ1]].\n6. Steps 3 and 5 give: $d(A, C)^2 = d(A, B)^2 + d(B, C)^2$ if and only if $\\angle ABC = \\pi / 2$.\n\n## Idea\n\nThe square of the side $AC$ is the sum of the squares of the other two sides, minus twice a dot product. This correction term vanishes exactly when the two sides at $B$ are perpendicular. So one computation gives the theorem and its converse.\n",
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