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  "body": "\n## Proof\n\nWe compute one dot product in two ways.\n\n1. Let $A$, $B$, $C$ be distinct points, and let $O$ be the [[midpoint|0S706JBZBG]] of $A$ and $C$, so $O = \\frac{1}{2}(A + C)$. Put $u = A - O$ and $z = O - B$.\n2. Then $C - O = -u$, so $A - B = u + z$ and $C - B = -u + z$.\n3. Hence the [[dot product|0S52TNNTH9]] is $(A - B) \\cdot (C - B) = (z + u) \\cdot (z - u) = \\norm{z}^2 - \\norm{u}^2 = \\norm{B - O}^2 - \\norm{A - O}^2$.\n4. The vectors $A - B$ and $C - B$ are nonzero, since $B \\neq A$ and $B \\neq C$. By the definition of the [[angle|0GJAJC6TNZ]] and [[the criterion for perpendicular vectors|03NK6RSRQ1]], $\\angle ABC = \\pi/2$ if and only if $(A - B) \\cdot (C - B) = 0$.\n5. By steps 3 and 4, $\\angle ABC = \\pi/2$ if and only if $\\norm{B - O} = \\norm{A - O}$.\n6. Direct statement: $A$, $B$, $C$ lie on a [[circle|060W5CBZP6]] whose centre is $O$. So $\\norm{B - O}$ and $\\norm{A - O}$ are both the radius, and step 5 gives $\\angle ABC = \\pi/2$.\n7. Converse: if $\\angle ABC = \\pi/2$, then step 5 gives $\\norm{B - O} = \\norm{A - O} = \\frac{1}{2}\\norm{A - C}$. This radius is positive since $A \\neq C$.\n8. The circle with the diameter $AC$ is the circle with the centre $O$ and the radius $\\frac{1}{2}\\norm{A - C}$. By step 7, $B$ lies on it.\n\n## Idea\n\nMeasured from the centre $O$, the two sides of the angle at $B$ are $z + u$ and $z - u$. Their dot product is the difference of the squares of $\\norm{z}$ and $\\norm{u}$. It vanishes exactly when $B$ is as far from $O$ as $A$ is.\n",
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