Question

 $Fie~H~ortocentrul~triunghiului~ABC~si~\overline{HM}=\frac{BC}{AH}\cdot\overline{AH},\\ \overline{HN}=\frac{AC}{BH}\cdot\overline{BH}~si~\overline{HP}=\frac{AB}{HC}\cdot\overline{HC}.Demonstrati~ca~\overline{HM}+\overline{HN}+\overline{HP}=\vec{0}$

Answer 1

$\displaystyle Folosind~faptul~ca~AH=2R \cos A~(vezi~demonstratia~in~privat) \\ \\ obtinem~\frac{BC}{AH}= \frac{a}{2R \cos A}= \frac{a}{\frac{a}{\sin A} \cdot \cos A}= \tan A. \\ \\ Analog~pentru~celelalte~rapoarte. \\ \\ Acum,~relatia~de~demonstrat~devine~\sum \tan A \cdot \overrightarrow{AH}= \overrightarrow{0}.$
$\displaystyle Dar~acest~lucru~este~cunoscut,~si~reiese~din~relatia \\ \\ \boxed{\overrightarrow{XH}= \frac{\tan A \cdot \overrightarrow{XA}+ \tan B \cdot \overrightarrow{XB}+ \tan C \cdot \overrightarrow{XC}}{\tan A+ \tan B+ \tan C}}~pentru~X=H.$