proof of pivot theorem

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




Combining this with A+B+C=π and EPF+FPD+DPE=2π, we get


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

Title proof of pivot theorem
Canonical name ProofOfPivotTheorem
Date of creation 2013-03-22 13:12:37
Last modified on 2013-03-22 13:12:37
Owner pbruin (1001)
Last modified by pbruin (1001)
Numerical id 4
Author pbruin (1001)
Entry type Proof
Classification msc 51M04