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  "body": "\n## Proof\n\nWe show that the two base angles have the same cosine.\n\n1. Let $b = d(A, C)$, $c = d(A, B)$ and $a = d(B, C)$, with the [[Euclidean distance|0NPPGHQNXQ]] $d$. By hypothesis $b = c$.\n2. The vertices of the [[triangle|0CTNFGYMM2]] are not collinear, so they are pairwise distinct. Thus $a, b, c \u003e 0$, and the [[angles|0GJAJC6TNZ]] at $B$ and $C$ are defined.\n3. For vectors $x, y \\in \\R^n$, the [[dot product|0S52TNNTH9]] is bilinear and symmetric. So $\\norm{x - y}^2 = \\norm{x}^2 - 2 \\, x \\cdot y + \\norm{y}^2$, with the [[length|0XKBKGTCPC]] $\\norm{\\cdot}$.\n4. With $x = A - B$ and $y = C - B$, we have $x - y = A - C$. Step 3 gives $b^2 = c^2 - 2 \\, x \\cdot y + a^2$. Since $b = c$, this is $(A - B) \\cdot (C - B) = a^2 / 2$.\n5. With $x = A - C$ and $y = B - C$, we have $x - y = A - B$. Step 3 gives $c^2 = b^2 - 2 \\, x \\cdot y + a^2$. Since $b = c$, this is $(A - C) \\cdot (B - C) = a^2 / 2$.\n6. By the definition of the angle and step 4, $\\cos \\angle ABC = \\dfrac{a^2 / 2}{c \\, a} = \\dfrac{a}{2c}$.\n7. By the definition of the angle and step 5, $\\cos \\angle ACB = \\dfrac{a^2 / 2}{b \\, a} = \\dfrac{a}{2b} = \\dfrac{a}{2c}$.\n8. Both angles lie in $[0, \\pi]$, and they have the same cosine. The cosine is injective on $[0, \\pi]$, by the uniqueness in the definition of the [[arccosine|0CWDVN132S]].\n9. Hence $\\angle ABC = \\angle ACB$.\n\n## Idea\n\nThe cosine of an angle of a triangle depends only on the three sides. At $B$ and at $C$ the formula uses the same numbers once $d(A, B) = d(A, C)$. Equal cosines in $[0, \\pi]$ mean equal angles.\n",
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