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

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