# Pivot Theorem

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

The Pivot Theorem:Let \(D\), \(E\), \(F\) be the points on the sides \(BC\), \(CA\), \(AB\) of \(\triangle ABC\), respectively. Then the circles \(AEF\), \(BDF\) and \(CDE\) pass through a common point, called the Miquel point. \(_\square\)

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 \(D\), \(E\), \(F\) 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 the 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 the points \(E, F\) and \(M\) must also pass through the point \(A,\) so that the quadrilateral \(AEMF\) is also cyclic, i.e. the 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 \(M\neq D\). So the 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 the quadrilateral \(AEMF\) is also cyclic, that is, the circle \(AEF\) passes through \(M\). So \(M\) is the Miquel point, and the proof is complete. \(_\square\)