Miquel's theorem

Statement

Let $ABC$ be a triangle in the Euclidean plane. Let $A'$ , $B'$ , $C'$ be points on the lines $BC$ , $CA$ , $AB$ , each different from the vertices of $ABC$ . Then the three circles through $A, B', C'$ , through $B, C', A'$ and through $C, A', B'$ have a common point.

Remarks

The three points on each circle are not collinear, since for example $B'$ and $C'$ lie on two different lines through $A$ and differ from $A$ . For a proof, let $M$ be the second common point of the first two circles, or their point of tangency. Chasing directed angles in the cyclic quadrilaterals $AB'MC'$ and $BC'MA'$ shows that $CA'MB'$ is cyclic too. The common point is called the Miquel point. Example: if $A'$ , $B'$ , $C'$ are the midpoints of the sides, the Miquel point is the circumcentre $O$ . Indeed, $\angle OB'A$ and $\angle OC'A$ are right angles, so $O$ lies on the circle with diameter $AO$ through $A$ , $B'$ , $C'$ .

Going deeper

For a student

Miquel's theorem is a standard exercise in angle chasing with cyclic quadrilaterals. It is the starting point of a family of results on Miquel points of complete quadrilaterals.

Reading path

To understand this page from the ground up, read these 3 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.

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Circle concept
Triangle concept
Miquel's theorem statement

Tag	`K8A688YSPW`
URI	`m2://K8A688YSPW`
Kind	statement, theorem
Writing	W0 stub
MSC	51M04
Status	This copy of the site has no build data. Run `m2 status K8A688YSPW`.
About	Triangle
Uses	Triangle, Circle
Same as	Wikidata Q2449984 Wikipedia Miquel's theorem