Miquel's Point Theorem

A proof that the circumcircles of the three sub-triangles formed by points on the sides of a triangle share a common point — the Miquel point.

Keywords: Miquel point, circumcircles, cyclic quadrilateral, concurrent circles

Prerequisites: cyclic-quadrilateral-proof · Difficulty: advanced

Let $D$, $E$, $F$ be points on sides $BC$, $CA$, $AB$ of triangle $ABC$. Then the circumcircles of triangles $AEF$, $BFD$, and $CDE$ all pass through a single common point $M$, the Miquel point.

Strategy: define $M$ as the second intersection of the circumcircles of $△AEF$ and $△BFD$. Inscribed-angle chasing at $M$ gives $\angle EMF=180°-\angle A$ and $\angle FMD=180°-\angle B$; the angles around $M$ then force $\angle EMD+\angle C=180°$, so $M$ also lies on the circumcircle of $△CDE$.

Step 1: Points on the Sides

Start with triangle $ABC$. Place points $D$ on $BC$, $E$ on $CA$, and $F$ on $AB$.

Step 2: Two Circumcircles Meet at $M$

Draw the circumcircle of $△AEF$ and the circumcircle of $△BFD$.

These two circles both pass through $F$. They must intersect at a second point as well — call it $M$.

Step 3: Three Circumcircles Share $M$

Miquel's Theorem. Let $D$, $E$, $F$ be points on sides $BC$, $CA$, $AB$ of $△ABC$. The circumcircles of $△AEF$, $△BFD$, and $△CDE$ all pass through a single common point $M$, called the Miquel point.

Proof strategy. $M$ is defined as the second intersection of the first two circumcircles. Angle chasing at $M$ on each of those circles gives $\angle EMF=180°-\angle A$ and $\angle FMD=180°-\angle B$. The three angles around $M$ sum to $360°$, so $\angle EMD=\angle A+\angle B=180°-\angle C$, placing $M$ on the circumcircle of $△CDE$.

Step 4: $M$ Lies on the Third Circumcircle

Angle chasing means deriving an angle relation by walking through the figure and applying the inscribed-angle theorem at each step. Here we use it at vertex $M$ twice.

On the circumcircle of $AEF$, $A$ and $M$ lie on opposite arcs of chord $EF$, so $\angle EAF+\angle EMF=180°$, giving $\angle EMF=180°-\angle A$.

Similarly on the circumcircle of $BFD$, $\angle FMD=180°-\angle B$.

The three angles at $M$ sum to $360°$, so $\angle EMD=360°-\angle EMF-\angle FMD=\angle A+\angle B$.

With $\angle A+\angle B+\angle C=180°$, this gives $\angle EMD+\angle C=180°$. Since $M$ and $C$ lie on opposite sides of $ED$, the four points $M$, $D$, $C$, $E$ are concyclic.

Step 5: Miquel's Point

All three circumcircles share the common point $M$ — the Miquel point of the configuration.

This result holds for any choice of points $D$, $E$, $F$ on the sides of the triangle.

The three circumcircles of triangles $AEF$, $BFD$, and $CDE$ all pass through the Miquel point $M$.

This elegant result generalizes: for any triangle and any points on its sides, these three circles always share exactly one common point. ∎