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  "body": "\n## Statement\n\nLet $ABC$ be a [[triangle|0CTNFGYMM2]] in the Euclidean plane.\nLet $A'$, $B'$, $C'$ be points on the lines $BC$, $CA$, $AB$, each different from the vertices of $ABC$.\nThen the three [[circles|060W5CBZP6]] through $A, B', C'$, through $B, C', A'$ and through $C, A', B'$ have a common point.\n\n## Remarks\n\nThe three points on each circle are not collinear, since for example $B'$ and $C'$ lie on two different lines through $A$ and differ from $A$.\nFor a proof, let $M$ be the second common point of the first two circles, or their point of tangency.\nChasing directed angles in the cyclic quadrilaterals $AB'MC'$ and $BC'MA'$ shows that $CA'MB'$ is cyclic too.\nThe common point is called the Miquel point.\nExample: if $A'$, $B'$, $C'$ are the midpoints of the sides, the Miquel point is the [[circumcentre|0F98FZK14A]] $O$.\nIndeed, $\\angle OB'A$ and $\\angle OC'A$ are right angles, so $O$ lies on the circle with diameter $AO$ through $A$, $B'$, $C'$.\n\n## Explanations\n\n### L1 human\n\nMark one point on each side of a triangle.\nEach corner and the two marked points next to it determine a circle.\nThe three circles you get always pass through one common point.\n\n### L2 human\n\nMiquel's theorem is a standard exercise in angle chasing with cyclic quadrilaterals.\nIt is the starting point of a family of results on Miquel points of complete quadrilaterals.\n",
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