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$P$ 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 $$ \frac{PD'}{AD}+\frac{PE'}{BE}+\frac{PZ'}{CZ}=1.$$

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I believe it solved using similar triangles and Thales's theorem, but it didn't lead me anywhere.

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

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