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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 $\angle AOB' = \angle AOI$

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2 Answers

up vote 1 down vote accepted

I got the answer! Let $OG$ intersect $BC$ at $L$ (I forgot to name the point). Let $\angle B = b, \angle EBB' = a$.

$\angle EBB' = \angle OLB' = a \implies \angle LOB' = 90 - a$

Similarly, $\angle AOG = 90 - (b - a)$

$\implies \angle AOB' = \angle AOG + \angle OLB' = 90 - a + 90 - (b - a) = 180 - b$ as desired.

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Alternative solution could be as follows:

  • Denote $\angle X = \angle OAC$, $\angle Y = \angle BAH$, then $\angle HAO = \angle A + \angle X + \angle Y$.
  • $\triangle ACD$ is right, so $Y = B-90^\circ$.
  • $\triangle AOB$ and $\triangle BOC$ are isosceles, so $(A+X) + (C+X) = B$, hence $X = B-90^\circ$.
  • Finally, $X+Y = 2B-180 = 2B - (A+B+C)$, therefore $$\angle HAO = \angle X+\angle Y+\angle A= \angle B-\angle C.$$

I hope this helps ;-)

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