proof of pivot theorem

Let $\mathrm{△}ABC$ be a triangle, and let $D$, $E$, and $F$ be pointson $BC$, $CA$, and $AB$, respectively. The circumcircles of$\mathrm{△}AEF$ and $\mathrm{△}BFD$ intersect in $F$ and in anotherpoint, which we call $P$. Then $AEPF$ and $BFPD$ are cyclicquadrilaterals, so

$$\mathrm{\angle }A+\mathrm{\angle }EPF=\pi$$

and

$$\mathrm{\angle }B+\mathrm{\angle }FPD=\pi$$

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

$$\mathrm{\angle }C+\mathrm{\angle }DPE=\pi .$$

This implies that $CDPE$ is a cyclic quadrilateral as well, so that$P$ lies on the circumcircle of $\mathrm{△}CDE$. Therefore, thecircumcircles of the triangles $AEF$, $BFD$, and $CDE$ have a commonpoint, $P$.