# proof of pivot theorem

Let $\triangle ABC$ be a triangle, and let $D$, $E$, and $F$ be points on $BC$, $CA$, and $AB$, respectively. The circumcircles of $\triangle AEF$ and $\triangle BFD$ intersect in $F$ and in another point, which we call $P$. Then $AEPF$ and $BFPD$ are cyclic quadrilaterals, so

 $\angle A+\angle EPF=\pi$

and

 $\angle B+\angle FPD=\pi$

Combining this with $\angle A+\angle B+\angle C=\pi$ and $\angle EPF+\angle FPD+\angle DPE=2\pi$, we get

 $\angle C+\angle DPE=\pi.$

This implies that $CDPE$ is a cyclic quadrilateral as well, so that $P$ lies on the circumcircle of $\triangle CDE$. Therefore, the circumcircles of the triangles $AEF$, $BFD$, and $CDE$ have a common point, $P$.

Title proof of pivot theorem ProofOfPivotTheorem 2013-03-22 13:12:37 2013-03-22 13:12:37 pbruin (1001) pbruin (1001) 4 pbruin (1001) Proof msc 51M04