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  "body": "\n## Proof\n\nWe write the two sides from $A$ as a sum and a difference of two vectors.\n\n1. By the definition of the [[midpoint|0S706JBZBG]], $D = B + \\frac{1}{2}(C - B)$.\n2. Let $m = D - A$ and $w = B - D$. By step 1, $w = -\\frac{1}{2}(C - B)$ and $C - D = \\frac{1}{2}(C - B) = -w$.\n3. Then $B - A = (D - A) + (B - D) = m + w$ and $C - A = (D - A) + (C - D) = m - w$.\n4. With the [[Euclidean distance|0NPPGHQNXQ]] and the [[length|0XKBKGTCPC]], $\\abs{AB} = \\norm{m + w}$, $\\abs{AC} = \\norm{m - w}$, $\\abs{AD} = \\norm{m}$ and $\\abs{BD} = \\norm{w}$.\n5. $\\R^n$ with the dot product is an inner product space with the norm $\\norm{\\cdot}$. By the [[parallelogram law|026HPGYGS4]], $\\norm{m + w}^2 + \\norm{m - w}^2 = 2 \\norm{m}^2 + 2 \\norm{w}^2$.\n6. By step 4, this is $\\abs{AB}^2 + \\abs{AC}^2 = 2 \\left( \\abs{AD}^2 + \\abs{BD}^2 \\right)$.\n\n## Idea\n\nSeen from $A$, the vertices $B$ and $C$ sit at $m + w$ and $m - w$. Here $m$ goes to the midpoint and $w$ is half the side $BC$. The parallelogram law for $m$ and $w$ is then exactly the theorem.\n",
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