Hey i have this problem:
Points D,E and F lie on the sides AB, BC and CA, respectively of triangle ABC.
Prove that the circumcircles of triangles ADF, BED and CFE meet in a common point.
Thanks
Let M be the point of intersection of the circumcircles of the triangles BDE and EFC. Then,
$\displaystyle \widehat{DME}=180-B$
$\displaystyle \widehat{EMF}=180-C$
$\displaystyle \widehat{DMF}=360-(180-B)-(180-C)=B+C=180-A$
That means M is on the circumcircle of the triangle ADF.