Consider the following triangle with orthocentre $H$ and circumcentre $O$. Prove that $\angle HAO = \angle B - \angle C$.
I am familiar with the proof for this when $ABC$ is acute, I wanted to prove it when it is obtuse.
$\angle HAO = \angle HAC + \angle CAO$
Consider $\Delta DAC$, clearly $\angle HAC = 90 - \angle C$. It remains to prove that $\angle CAO = \angle B - 90$. Or, $\angle AOI = 180 - \angle B = \angle A + \angle C$. This is where I'm stuck.
EDIT: I realized that in quadrilateral $AOB'B$, $\angle AOB' = 180 - \angle B$, so the proof reduces to proving that $\angle AOB' = \angle AOI$