Proof of side-side-side congruence with orthonormal frames

Proof

The direction "only if" is immediate. For "if", we build the isometry from two orthonormal frames.

1. Suppose that the triangles are congruent. Then there is an isometry $f$ of $\R^n$ with $f(A) = A'$, $f(B) = B'$ and $f(C) = C'$.
2. 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".
3. For "if", assume the three equalities. Let $u = B - A$, $v = C - A$, $u' = B' - A'$ and $v' = C' - A'$. With the Euclidean distance and the length, the hypotheses say $\norm{u} = \norm{u'}$, $\norm{v} = \norm{v'}$ and $\norm{u - v} = \norm{u' - v'}$.
4. The dot product 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, $u \cdot v = u' \cdot v'$.
5. Every difference of two vertices of $ABC$ lies in the span of $u$ and $v$. Since $A$, $B$, $C$ are not collinear, this span has dimension $2$. So $u$ and $v$ are linearly independent, and $n \geq 2$. The same holds for $u'$ and $v'$.
6. 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.
7. 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.
8. 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 $\norm{h'}^2 = \norm{v'}^2 - \lambda'^2$.
9. 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 $\norm{h} = \norm{h'}$.
10. Each orthonormal family of $\R^n$ extends to an orthonormal basis. Let $(e_1, \dots, e_n)$ and $(e_1', \dots, e_n')$ be orthonormal bases that extend the two families.
11. Let $T$ be the linear map 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.
12. 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.
13. 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.
14. We have $f(A) = A'$. Also $f(B) = A' + T(u) = A' + \norm{u} \, e_1' = A' + \norm{u'} \, e_1' = A' + u' = B'$.
15. 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.
16. Thus $f(C) = A' + T(v) = A' + v' = C'$.
17. By steps 13, 14 and 16, $f$ is an isometry that maps $A$, $B$, $C$ to $A'$, $B'$, $C'$. So the triangles are congruent.