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  "body": "\n## Proof\n\nWe compute each sine from the three sides and find that the three quotients are given by one symmetric formula.\n\n1. The vertices of the [[triangle|0CTNFGYMM2]] are not [[collinear|0AFNNTRTJE]], so they are distinct. Thus $a, b, c \u003e 0$, and the three [[angles|0GJAJC6TNZ]] are defined.\n2. We apply the [[law of cosines|060ZVZBV3E]] with the vertices renamed, so that $A$ plays the role of $C$. It gives $a^2 = b^2 + c^2 - 2bc \\cos \\alpha$. So $\\cos \\alpha = \\dfrac{b^2 + c^2 - a^2}{2bc}$.\n3. In the same way, $\\cos \\beta = \\dfrac{c^2 + a^2 - b^2}{2ca}$ and $\\cos \\gamma = \\dfrac{a^2 + b^2 - c^2}{2ab}$.\n4. Let $Q = 4b^2c^2 - (b^2 + c^2 - a^2)^2$. Two differences of squares give $Q = \\left( a^2 - (b - c)^2 \\right) \\left( (b + c)^2 - a^2 \\right) = (a - b + c)(a + b - c)(b + c - a)(a + b + c)$.\n5. A permutation of $a$, $b$, $c$ only permutes the four factors in step 4. So step 4 with the letters renamed gives $Q = 4c^2a^2 - (c^2 + a^2 - b^2)^2 = 4a^2b^2 - (a^2 + b^2 - c^2)^2$.\n6. Let $x = B - A$ and $y = C - A$. With the [[length|0XKBKGTCPC]] and the [[dot product|0S52TNNTH9]], $\\norm{x} = c$, $\\norm{y} = b$ and $x \\cdot y = bc \\cos \\alpha = \\frac{1}{2}(b^2 + c^2 - a^2)$. So $Q = 4 \\left( \\norm{x}^2 \\norm{y}^2 - (x \\cdot y)^2 \\right)$.\n7. Every difference of two vertices lies in the span of $x$ and $y$. 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. The [[sine|08WFZP6P89]] is nonnegative on $[0, \\pi]$, and $\\alpha \\in [0, \\pi]$. By the [[identity $\\sin^2 + \\cos^2 = 1$|03H7WZ4T3K]] and step 2, $\\sin^2 \\alpha = 1 - \\cos^2 \\alpha = \\dfrac{Q}{4b^2c^2}$. So $\\sin \\alpha = \\dfrac{R}{2bc} \u003e 0$.\n10. In the same way, steps 3 and 5 give $\\sin \\beta = \\dfrac{R}{2ca} \u003e 0$ and $\\sin \\gamma = \\dfrac{R}{2ab} \u003e 0$. So the three quotients are defined.\n11. By steps 9 and 10, $\\dfrac{a}{\\sin \\alpha} = \\dfrac{2abc}{R}$, $\\dfrac{b}{\\sin \\beta} = \\dfrac{2abc}{R}$ and $\\dfrac{c}{\\sin \\gamma} = \\dfrac{2abc}{R}$.\n12. Hence $\\dfrac{a}{\\sin \\alpha} = \\dfrac{b}{\\sin \\beta} = \\dfrac{c}{\\sin \\gamma}$.\n\n## Idea\n\nThe law of cosines gives each cosine from the sides, and then each sine. All three sines have the same numerator $\\sqrt{Q}$, a symmetric expression in the sides, which is four times the area. Dividing a side by the sine of the opposite angle then gives $2abc / \\sqrt{Q}$ every time.\n",
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