Radii of the circles circumscribing the triangles $BHC,CHA,AHB,ABC$ are all equal

Question

If $ H$ is the orthocenter of a triangle $ABC$;prove that the radii of the circles circumscribing the triangles $BHC,CHA,AHB,ABC$ are all equal.

Answer 1

Since $\widehat{AHB}+\widehat{ACB}=\pi$, $AHB$ and $ABC$ have the same circumradius by the sine theorem, since they have $AB$ in common and $\sin\widehat{AHB}=\sin\widehat{ACB}$.

Answer 2

Hint: Let $H_A$, $H_B$, and $H_C$ be the reflections of $H$ about $BC$, $CA$, and $AB$, respectively. Prove that $H_A$, $H_B$, and $H_C$ are on the circumscribed circle of $ABC$.