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  "body": "\n## Proof\n\nWe expand the square of the length of a difference of two vectors.\n\n1. Let $u = A - C$ and $v = B - C$. With the [[Euclidean distance|0NPPGHQNXQ]] and the [[length|0XKBKGTCPC]], $\\norm{u} = b$ and $\\norm{v} = a$.\n2. We have $u - v = A - B$, so $c^2 = \\norm{u - v}^2 = (u - v) \\cdot (u - v)$, with the [[dot product|0S52TNNTH9]].\n3. The dot product is bilinear and symmetric. So $(u - v) \\cdot (u - v) = u \\cdot u - 2 \\, u \\cdot v + v \\cdot v = b^2 + a^2 - 2 \\, u \\cdot v$.\n4. The vertices of the [[triangle|0CTNFGYMM2]] are not collinear, so they are distinct. Thus $u \\neq 0$ and $v \\neq 0$, and the [[angle|0GJAJC6TNZ]] $\\gamma$ at $C$ is defined.\n5. By the definition of the angle at $C$, $\\cos \\gamma = \\dfrac{u \\cdot v}{\\norm{u} \\, \\norm{v}} = \\dfrac{u \\cdot v}{ab}$. So $u \\cdot v = ab \\cos \\gamma$.\n6. Steps 2, 3 and 5 give $c^2 = a^2 + b^2 - 2ab \\cos \\gamma$.\n\n## Idea\n\nPut the vertex $C$ at the origin. The side $AB$ is the difference of the two vectors from $C$. Squaring its length produces the two squared sides and a cross term, and the cross term is the dot product, which is $ab \\cos \\gamma$.\n",
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