Euler line proof

Let $O$ the circumcenter of $\\triangle ABC$ and $G$ its centroid. Extend $O G$ until a point $P$ such that $OG/GP=1/2$ . We’ll prove that $P$ is the orthocenter $H$ .

Draw the median $AA^{\\prime}$ where $A^{\\prime}$ is the midpoint of $B C$ . Triangles $OGA^{\\prime}$ and $P G A$ are similar, since $GP=2GO$ , $AG=2A^{\\prime}G$ and $\\angle OGA^{\\prime}=\\angle PGA$ . Then $\\angle OA^{\\prime}G=\\angle PGA$ and $OA^{\\prime}\\parallel AP$ . But $OA^{\\prime}\\perp BC$ so $AP\\perp BC$ , that is, $A P$ is a height of the triangle.

Repeating the same argument for the other medians proves that $P$ lies on the three heights and therefore it must be the orthocenter $H$ .

The ratio is $OG/GH=1/2$ since we constructed it that way.

单词	EulerLineProof
释义	Euler line proof Let $O$ the circumcenter of $\\triangle ABC$ and $G$ its centroid. Extend $O G$ until a point $P$ such that $OG/GP=1/2$ . We’ll prove that $P$ is the orthocenter $H$ . Draw the median $AA^{\\prime}$ where $A^{\\prime}$ is the midpoint of $B C$ . Triangles $OGA^{\\prime}$ and $P G A$ are similar, since $GP=2GO$ , $AG=2A^{\\prime}G$ and $\\angle OGA^{\\prime}=\\angle PGA$ . Then $\\angle OA^{\\prime}G=\\angle PGA$ and $OA^{\\prime}\\parallel AP$ . But $OA^{\\prime}\\perp BC$ so $AP\\perp BC$ , that is, $A P$ is a height of the triangle. Repeating the same argument for the other medians proves that $P$ lies on the three heights and therefore it must be the orthocenter $H$ . The ratio is $OG/GH=1/2$ since we constructed it that way.
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