# 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:

## Statement

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.

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.

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.

## Proof

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. \(_\square\)

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. \(_\square\)