Pivot Theorem

The Pivot theorem is a special case of Miquel's theorem. It is trivial to prove but is a really surprising result. It states the following:

Let $A'$, $B'$, $C'$ be the points on sides $BC, CA, AB$ of $\triangle ABC$, respectively. Then circles $AB'C', A'BC',$ and $A'B'C$ pass through a common point called the Miquel point.

Miquel point $\color{#D61F06}{M}$ in Pivot theorem

Let $A'$, $B'$, $C'$ be the points on sides $BC$, $CA$, $AB$ of $\triangle ABC$ (or even their extensions), respectively. The Miquel point lies on the circumcircle of $\triangle ABC$ if and only if points $A', B', C'$ are collinear.

Miquel point $\color{#D61F06}{M}$

If you have difficulties in identifying the circles, take the circumcircle of a vertex and the two chosen points lying on the two sides through (adjacent to) that vertex.

The Pivot theorem is the case when $D$, $E$, $F$ $(A', B', C',$ respectively in the theorem above$)$ are not collinear. This result can be proved only by chasing a few angles as follows:

Consider the above configuration. Suppose that circles $BDF$ and $CDE$ intersect at $M\neq D$. So quadrilaterals $BDMF$ and $CEMD$ are cyclic, meaning

$$\angle DMF=180^\circ-\angle B\ \ \text{ and }\ \ \angle DME=180^\circ-\angle C,$$

giving $\angle EMF=360^\circ-(\angle DMF+\angle DME)=\angle B+\angle C.$ Since $\angle B+\angle C$ also equals $180^\circ-\angle A$ from $\triangle ABC,$ the circle passing through points $E, F, M$ must also pass through point $A,$ so that quadrilateral $AEMF$ is also cyclic, i.e. circle $AEF$ passes through $M$. So, $M$ is the Miquel point, and the proof is complete. $_\square$

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Consider the above configuration. To chase angles, we'll use directed angles $\bmod$ $\pi$ to avoid configuration issues. Suppose that circles $BDF$ and $CDE$ intersect at $M\neq D$. Then quadrilaterals $BDMF$ and $CEMD$ are cyclic, meaning $\measuredangle MFB=\measuredangle MDB$ and $\measuredangle MDC=\measuredangle MEC$. Now just track that

$$\measuredangle MFA=\measuredangle MFB=\measuredangle MDB=\measuredangle MDC=\measuredangle MEC=\measuredangle MEA.$$

So quadrilateral $AEMF$ is also cyclic, that is, circle $AEF$ passes through $M$. So $M$ is the Miquel point, and the proof is complete. $_\square$