Let $H$ be the orthocenter in a triangle with sides $a, b, c$. Is it true that $$a^2 + HA^2 = 4R^2$$ where $R$ is the circumradius?
2026-05-05 01:30:09.1777944609
Formula in a triangle
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We consider $B'$reflection of $H$ versus midpoint of $AC$. Because $AHCB'$is parallelogram results that $CB'$ parallel with $AH$ and has the same length. Consequently the triangle $BCB'$ is rectangle and $BB'$ is the diameter of the circumscribed circle. Clear $BC^2+ HA^2=BB'^2$ and $ a^2 + HA^2 =4R^2$.