# $P$ is a point inside the triangle $ABC$, from which perpendiculars intersect the sides $BC$ $AC$ $AB$ to the points…

$$P$$ 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.

• Hint: $$\frac{|PD'|}{|AD|}= \frac{\frac12|BC|\cdot|PD'|}{\frac12|BC|\cdot|AD|}=\frac{|\triangle PBC|}{|\triangle ABC|}$$ – Blue Sep 21 '18 at 19:21
• Thank you! I didn't think of that! – zevs12 Sep 22 '18 at 9:57