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

Miquel's Point Theorem

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

Proof Strategy

  1. The circumcircles of $\triangle AEF$ and $\triangle BFD$ intersect at $F$ and at some second point $M$.
  2. In the circumcircle of $AEF$: $\angle AMF = \angle AEF$ (inscribed angles on the same arc).
  3. In the circumcircle of $BFD$: $\angle BMD = \angle BFD$ (inscribed angles on the same arc).
  4. Angle chasing shows that $\angle DME + \angle DCE = 180°$, so $M$ lies on the circumcircle of $\triangle CDE$.

The Proof

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 $\triangle AEF$ and the circumcircle of $\triangle BFD$.

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 $\triangle ABC$. The circumcircles of $\triangle AEF$, $\triangle BFD$, and $\triangle CDE$ 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 $\triangle CDE$.

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.

Conclusion

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.