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  "body": "\n## Statement\n\nLet $PQ$ be a chord of a [[circle|060W5CBZP6]] in the Euclidean plane and let $M$ be its [[midpoint|0S706JBZBG]].\nLet $AB$ and $CD$ be two further chords through $M$.\nSuppose that the segment $AD$ meets $PQ$ at $X$ and the segment $BC$ meets $PQ$ at $Y$.\nThen $M$ is the midpoint of $XY$.\n\n## Remarks\n\nOne proof drops perpendiculars from $X$ and $Y$ to the chords and uses similar triangles together with the power of the point $M$, $MA \\cdot MB = MC \\cdot MD = MP \\cdot MQ$.\nA projective proof uses the involution that the pencil of conics through $A, B, C, D$ induces on the line $PQ$.\nExample: let $CD$ be the mirror image of $AB$ in the diameter through $M$ perpendicular to $PQ$, with $C$ the image of $A$.\nThen the figure is symmetric and $MX = MY$ is clear.\nThe theorem stays true when the circle is replaced by any conic.\n\n## Explanations\n\n### L1 human\n\nTake a chord of a circle and its midpoint.\nThrough the midpoint draw two more chords, and join their ends crosswise to make a shape like the wings of a butterfly.\nThe wings cut the first chord at two points that are exactly equally far from the midpoint.\n",
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