ISL 1979 Problem 64

From point $P$ on arc $BC$ of the circumcircle about triangle $ABC$ , $PX$ is constructed perpendicular to $BC$ , $PY$ is perpendicular to $AC$ , and $PZ$ perpendicular to $AB$ (all extended if necessary). Prove that $\frac{BC}{PX}=\frac{AC}{PY}+\frac{AB}{PZ}$ .