Property of angle bisectors in a triangle Let $ABC$ be a triangle having circumcenter $O$. Suppose $AH$ is the altitude from vertex $A$ and $AT$ bisects angle $A$. I would like a simple geometric proof that $AT$ also bisects angle $OAH$.

Without loss of generality, assume that $\angle B \geq \angle C$. Now suppose that $\angle B \leq \pi / 2$. Drop a perpendicular from $O$ to $D$ on $\overline{AC}$.

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By the Inscribed Angle Theorem, $\angle AOC = 2 \angle B$, so $\angle AOD = \angle B$. Therefore, $\angle BAH$ and $\angle OAD$ are congruent complements of $\angle B$, whereupon $$\angle HAT = \angle BAT - \angle BAH = \angle CAT - \angle OAD = \angle OAT$$ so that $\overline{AT}$ bisects $\angle OAH$.

Now suppose that $\angle B > \pi / 2$.

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We have $\angle AOC + 2\angle B = 2\pi$, whence $\angle AOD = \pi - \angle B$. Therefore, $\angle BAH$ and $\angle OAD$ are congruent complements of $\angle AOD$, whereupon $$\angle HAT = \angle BAT + \angle BAH= \angle CAT + \angle OAD = \angle OAT.$$