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  "body": "\n## Statement\n\nLet $ABCD$ be a simple quadrilateral, that is a simple [[polygon|8VEFWD3EJB]] with four vertices, in the Euclidean plane.\nLet $K$, $L$, $M$, $N$ be the [[midpoints|0S706JBZBG]] of $AB$, $BC$, $CD$, $DA$.\nThen $KLMN$ is a parallelogram.\nIts sides are parallel to the diagonals $AC$ and $BD$ and half as long, and its area is half the area of $ABCD$.\n\n## Remarks\n\nWith vectors, $L - K = \\frac{1}{2}(C - A) = M - N$, so $KL$ and $NM$ are parallel and equal, which makes $KLMN$ a parallelogram.\nIn the same way $KN$ and $LM$ equal $\\frac{1}{2}(D - B)$.\nThe area of $ABCD$ is $\\frac{1}{2}\\abs{(C - A) \\times (D - B)}$, and the area of $KLMN$ is $\\abs{\\frac{1}{2}(C - A) \\times \\frac{1}{2}(D - B)}$, which is half of it.\nExample: for the square with vertices $(0, 0), (2, 0), (2, 2), (0, 2)$, the midpoints form a square of side $\\sqrt 2$ and area $2$, half of $4$.\nThe parallelogram part needs no simplicity; for a self-crossing quadrilateral the area relation fails, and the parallelogram can even degenerate.\n\n## Explanations\n\n### L1 human\n\nDraw any four-sided figure and mark the middle of each side.\nJoining these four midpoints always gives a parallelogram.\nThis works even for very irregular quadrilaterals, and the parallelogram covers exactly half of the area.\n\n### L2 human\n\nThe proof is the midpoint theorem for triangles applied twice, once in each triangle cut off by a diagonal.\nIt is a good first example of how vectors turn a geometric statement into one line of algebra.\n",
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