$P$ is a point inside the triangle $ABC$, from which perpendiculars intersect the sides $BC$ $AC$ $AB$ to the points $D'$, $E'$, $Z'$. If $AD$, $BE$, $CZ$ are the heights prove that $$ \frac{PD'}{AD}+\frac{PE'}{BE}+\frac{PZ'}{CZ}=1.$$
I believe it solved using similar triangles and Thales's theorem, but it didn't lead me anywhere.