Consider triangle $ABC$. Let point $H$ be the orthocentre and point $O$ be the circumcentre.
Prove that area of one of triangles $AOH$,$BOH$,$COH$ is equal to the sum of areas of other two.
2026-03-27 22:52:55.1774651975
Area combo of inner triangles formed by orthocenter, circumcenter, and vertices
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Hint 1. $\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$.
Hint 2. Area of $\triangle AOH=\frac12 |\overrightarrow{OA}\times\overrightarrow{OH}|$.