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Let $\mathrm{H}$ be the orthocenter of an acute - angled triangle $\mathrm{ABC}$ and $\mathrm{O}$ be its circumcenter. Then $\overrightarrow{\mathrm{HA}}+\overrightarrow{\mathrm{HB}}+\overrightarrow{\mathrm{HC}}$
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Verified Answer
The correct answer is:
is equal to $2 \overrightarrow{\mathrm{HO}}$
G is centroid
$\begin{array}{l}
\mathrm{G}=\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}}{3} \\
\mathrm{G}=\frac{20+\mathrm{H}}{3} \\
2 \mathrm{O}+\mathrm{H}=3 \mathrm{G} \\
\overrightarrow{\mathrm{HA}}+\overrightarrow{\mathrm{HB}}+\overrightarrow{\mathrm{HC}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{H}}+\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{H}}+\overrightarrow{\mathrm{C}}-\overrightarrow{\mathrm{H}} \\
\quad=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}-3 \overrightarrow{\mathrm{H}} \\
\quad=3 \overrightarrow{\mathrm{G}}-3 \overrightarrow{\mathrm{H}} \\
\quad=2 \overrightarrow{\mathrm{O}}+\overrightarrow{\mathrm{H}}-3 \overrightarrow{\mathrm{H}} \\
\quad=2 \overrightarrow{\mathrm{O}}-2 \overrightarrow{\mathrm{H}} \\
\quad=2 \overrightarrow{\mathrm{HO}}
\end{array}$
$\begin{array}{l}
\mathrm{G}=\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}}{3} \\
\mathrm{G}=\frac{20+\mathrm{H}}{3} \\
2 \mathrm{O}+\mathrm{H}=3 \mathrm{G} \\
\overrightarrow{\mathrm{HA}}+\overrightarrow{\mathrm{HB}}+\overrightarrow{\mathrm{HC}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{H}}+\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{H}}+\overrightarrow{\mathrm{C}}-\overrightarrow{\mathrm{H}} \\
\quad=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}-3 \overrightarrow{\mathrm{H}} \\
\quad=3 \overrightarrow{\mathrm{G}}-3 \overrightarrow{\mathrm{H}} \\
\quad=2 \overrightarrow{\mathrm{O}}+\overrightarrow{\mathrm{H}}-3 \overrightarrow{\mathrm{H}} \\
\quad=2 \overrightarrow{\mathrm{O}}-2 \overrightarrow{\mathrm{H}} \\
\quad=2 \overrightarrow{\mathrm{HO}}
\end{array}$
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