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  "body": "\n## Proof\n\nWe compute the cosine and the sine of $\\alpha + \\beta$ from the side lengths, and then compare with $\\pi - \\gamma$.\n\n1. Let $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$ and $\\gamma = \\angle ACB$. Let $a = d(B, C)$, $b = d(C, A)$ and $c = d(A, B)$, with the [[Euclidean distance|0NPPGHQNXQ]] $d$.\n2. The vertices of a [[triangle|0CTNFGYMM2]] are not [[collinear|0AFNNTRTJE]]. So they are pairwise distinct, and $a, b, c \u003e 0$. Thus the three [[angles|0GJAJC6TNZ]] 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. Let $x = B - A$ and $y = C - A$. Then $x - y = B - C$, and step 3 gives $a^2 = c^2 - 2 \\, x \\cdot y + b^2$. So $x \\cdot y = \\frac{1}{2}(b^2 + c^2 - a^2)$.\n5. By the definition of the angle at $A$, $\\cos \\alpha = \\dfrac{x \\cdot y}{bc} = \\dfrac{b^2 + c^2 - a^2}{2bc}$. The same computation at $B$ and at $C$ gives $\\cos \\beta = \\dfrac{c^2 + a^2 - b^2}{2ca}$ and $\\cos \\gamma = \\dfrac{a^2 + b^2 - c^2}{2ab}$.\n6. Let $Q = 4b^2c^2 - (b^2 + c^2 - a^2)^2$. By step 4, $Q = 4 \\left( \\norm{x}^2 \\norm{y}^2 - (x \\cdot y)^2 \\right)$.\n7. Every difference of two vertices lies in the span of $x$ and $y$; for example $C - B = y - x$. Since the vertices are not collinear, this span has dimension $2$. So $x$ and $y$ are [[linearly independent|0CRMGPAH83]].\n8. By the [[Cauchy–Schwarz inequality|0RFWW56Y22]] and its case of equality, $(x \\cdot y)^2 \u003c \\norm{x}^2 \\norm{y}^2$. So $Q \u003e 0$. Let $R = \\sqrt{Q} \u003e 0$.\n9. Two differences of squares give $Q = \\left( a^2 - (b - c)^2 \\right) \\left( (b + c)^2 - a^2 \\right) = (a - b + c)(a + b - c)(b + c - a)(a + b + c)$.\n10. A permutation of $a$, $b$, $c$ only permutes the four factors in step 9. So the computation of step 9 with the letters renamed also gives $Q = 4c^2a^2 - (c^2 + a^2 - b^2)^2 = 4a^2b^2 - (a^2 + b^2 - c^2)^2$.\n11. The [[sine|08WFZP6P89]] is nonnegative on $[0, \\pi]$, and $\\alpha \\in [0, \\pi]$. By the [[identity $\\sin^2 + \\cos^2 = 1$|03H7WZ4T3K]] and step 5, $\\sin^2 \\alpha = 1 - \\cos^2 \\alpha = \\dfrac{Q}{4b^2c^2}$. So $\\sin \\alpha = \\dfrac{R}{2bc}$.\n12. In the same way, steps 5 and 10 give $\\sin \\beta = \\dfrac{R}{2ca}$ and $\\sin \\gamma = \\dfrac{R}{2ab}$. All three sines are positive.\n13. By the [[addition formula|028XEQCE88]], $\\cos(\\alpha + \\beta) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta = \\dfrac{(b^2 + c^2 - a^2)(c^2 + a^2 - b^2) - Q}{4abc^2}$.\n14. We insert $Q = 4b^2c^2 - (b^2 + c^2 - a^2)^2$ in the numerator of step 13. The numerator becomes $(b^2 + c^2 - a^2) \\cdot 2c^2 - 4b^2c^2 = 2c^2 (c^2 - a^2 - b^2)$.\n15. So $\\cos(\\alpha + \\beta) = \\dfrac{c^2 - a^2 - b^2}{2ab} = -\\cos \\gamma$, by step 5.\n16. By the addition formula, $\\sin(\\alpha + \\beta) = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta = \\dfrac{R \\left( (c^2 + a^2 - b^2) + (b^2 + c^2 - a^2) \\right)}{4abc^2}$.\n17. The sum in the numerator of step 16 is $2c^2$. So $\\sin(\\alpha + \\beta) = \\dfrac{R}{2ab} = \\sin \\gamma \u003e 0$.\n18. By [[$\\sin \\pi = 0$|01A8JGR4JM]] and $\\sin 0 = 0$, step 12 shows that no angle is $0$ or $\\pi$. So $\\alpha, \\beta \\in (0, \\pi)$, and $0 \u003c \\alpha + \\beta \u003c 2\\pi$.\n19. Suppose that $\\alpha + \\beta \u003e \\pi$, and let $z = \\alpha + \\beta - \\pi$. Then $z \\in (0, \\pi)$, so $\\sin z \\geq 0$.\n20. By the addition formula and [[$\\cos \\pi = -1$, $\\sin \\pi = 0$|01A8JGR4JM]], $\\sin(\\alpha + \\beta) = \\sin(z + \\pi) = -\\sin z \\leq 0$. This contradicts step 17. So $\\alpha + \\beta \\leq \\pi$.\n21. By the addition formula, $\\cos(\\pi - \\gamma) = \\cos \\pi \\cos(-\\gamma) - \\sin \\pi \\sin(-\\gamma) = -\\cos \\gamma$. Here we use that the cosine is even.\n22. Both $\\alpha + \\beta$ and $\\pi - \\gamma$ lie in $[0, \\pi]$, and by steps 15 and 21 they have the same cosine.\n23. The cosine is injective on $[0, \\pi]$, by the uniqueness in the definition of the [[arccosine|0CWDVN132S]]. So $\\alpha + \\beta = \\pi - \\gamma$.\n24. Hence $\\angle BAC + \\angle CBA + \\angle ACB = \\pi$.\n\n## Idea\n\nThe law of cosines gives the cosine of each angle from the three sides, and then the sine as well. With the addition formulas, $\\alpha + \\beta$ has the same cosine and sine as $\\pi - \\gamma$. The sine condition rules out $\\alpha + \\beta \u003e \\pi$, and on $[0, \\pi]$ the cosine determines the angle.\n",
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