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  "body": "\n## Proof\n\nThe direction \"only if\" is immediate. For \"if\", we build the isometry from two orthonormal frames.\n\n1. Suppose that the triangles are [[congruent|093EDENVA1]]. Then there is an [[isometry|02HPP5VCZH]] $f$ of $\\R^n$ with $f(A) = A'$, $f(B) = B'$ and $f(C) = C'$.\n2. So $d(A', B') = d(f(A), f(B)) = d(A, B)$, and in the same way $d(B', C') = d(B, C)$ and $d(C', A') = d(C, A)$. This proves \"only if\".\n3. For \"if\", assume the three equalities. Let $u = B - A$, $v = C - A$, $u' = B' - A'$ and $v' = C' - A'$. With the [[Euclidean distance|0NPPGHQNXQ]] and the [[length|0XKBKGTCPC]], the hypotheses say $\\norm{u} = \\norm{u'}$, $\\norm{v} = \\norm{v'}$ and $\\norm{u - v} = \\norm{u' - v'}$.\n4. The [[dot product|0S52TNNTH9]] is bilinear and symmetric, so $u \\cdot v = \\frac{1}{2} \\left( \\norm{u}^2 + \\norm{v}^2 - \\norm{u - v}^2 \\right)$. The same holds for $u'$ and $v'$. By step 3, $u \\cdot v = u' \\cdot v'$.\n5. Every difference of two vertices of $ABC$ lies in the span of $u$ and $v$. Since $A$, $B$, $C$ are not [[collinear|0AFNNTRTJE]], this span has dimension $2$. So $u$ and $v$ are [[linearly independent|0CRMGPAH83]], and $n \\geq 2$. The same holds for $u'$ and $v'$.\n6. Let $e_1 = u / \\norm{u}$, let $\\lambda = v \\cdot e_1$, and let $h = v - \\lambda e_1$. Then $h \\neq 0$, since otherwise $v$ is a multiple of $u$, against step 5.\n7. We have $e_1 \\cdot h = \\lambda - \\lambda = 0$ and $\\norm{h}^2 = \\norm{v}^2 - 2 \\lambda^2 + \\lambda^2 = \\norm{v}^2 - \\lambda^2$. Let $e_2 = h / \\norm{h}$. Then $(e_1, e_2)$ is an [[orthonormal family|0RW9TNNT87]].\n8. In the same way let $e_1' = u' / \\norm{u'}$, $\\lambda' = v' \\cdot e_1'$, $h' = v' - \\lambda' e_1'$ and $e_2' = h' / \\norm{h'}$. Then $(e_1', e_2')$ is an orthonormal family and $\\norm{h'}^2 = \\norm{v'}^2 - \\lambda'^2$.\n9. By steps 3 and 4, $\\lambda = (u \\cdot v) / \\norm{u} = (u' \\cdot v') / \\norm{u'} = \\lambda'$. With step 3 and steps 7 and 8, also $\\norm{h} = \\norm{h'}$.\n10. Each orthonormal family of $\\R^n$ [[extends to an orthonormal basis|0Y85GYEGSF]]. Let $(e_1, \\dots, e_n)$ and $(e_1', \\dots, e_n')$ be [[orthonormal bases|0S9GGTR0EX]] that extend the two families.\n11. Let $T$ be the [[linear map|0NRYHJPQG2]] of $\\R^n$ with $T(e_i) = e_i'$ for each $i$. It exists because a linear map can take any values on a basis.\n12. For $x = \\sum_i x_i e_i$, orthonormality gives $\\norm{T(x)}^2 = \\norm{\\sum_i x_i e_i'}^2 = \\sum_i x_i^2 = \\norm{x}^2$. So $T$ preserves lengths.\n13. Let $f(x) = A' + T(x - A)$. By linearity, $f(x) - f(y) = T(x - y)$. By step 12, $d(f(x), f(y)) = d(x, y)$ for all $x, y$. So $f$ is an isometry.\n14. We have $f(A) = A'$. Also $f(B) = A' + T(u) = A' + \\norm{u} \\, e_1' = A' + \\norm{u'} \\, e_1' = A' + u' = B'$.\n15. By steps 6 and 7, $v = \\lambda e_1 + \\norm{h} \\, e_2$. So $T(v) = \\lambda e_1' + \\norm{h} \\, e_2' = \\lambda' e_1' + \\norm{h'} \\, e_2' = \\lambda' e_1' + h' = v'$, by step 9.\n16. Thus $f(C) = A' + T(v) = A' + v' = C'$.\n17. By steps 13, 14 and 16, $f$ is an isometry that maps $A$, $B$, $C$ to $A'$, $B'$, $C'$. So the triangles are congruent.\n\n## Idea\n\nThe three side lengths fix the lengths of the edge vectors $u$, $v$ from $A$ and their dot product. These numbers fix the coordinates of $u$ and $v$ in an orthonormal frame adapted to them. A linear isometry that carries one frame to the other, followed by a translation, carries one triangle to the other.\n",
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