three circles

rogue

Let $P$ be arbitrary point inside the triangle $ABC,$ $\omega_A,$ $\omega_B$ and $\omega_C$ be the circumcircles of triangles $BPC,$ $APC$ and $APB$ respectively. Denote by $X,Y,Z$ the intersection points of straight lines $AP,$ $BP,$ $CP$ with circles $\omega_A,$ $\omega_B,$ $\omega_C$ respectively ( $X,Y,Z\ne P$ ). Prove that $\frac{AP}{AX}+\frac{BP}{BY}+\frac{CP}{CZ}=1.$

**Posted:** Sat May 24, 2008 1:51 am

mr.danh

This problem can be solved easily with Ptoleme theorem.

**Posted:** Sat May 24, 2008 2:30 am

The QuattoMaster 6000

Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1170198863&t=191082

**Posted:** Sat May 24, 2008 8:10 am