# 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 $\overline{AC}$.

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 $\angle B > \pi / 2$.

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 $$\angle HAT = \angle BAT + \angle BAH= \angle CAT + \angle OAD = \angle OAT.$$

• Nice approach, but I believe you want to drop a perpendicular from $O$ to $D$ on $\overline{AC}$. Also, I think you want $\angle HAT = \angle BAH - \angle BAT = \angle OAD - \angle CAT = \angle OAT$. Also, I believe the argument requires modification if $\angle B > \pi / 2$. Commented Jan 10, 2014 at 17:03
• Very good argument. I realized after my first comment that in the $O$ inside the triangle case, $\angle BAT$ may be larger or smaller than $\angle BAH$. Your proof doesn't cover the latter case. Commented Jan 10, 2014 at 20:21
• @RichardHevener: I believe your concern can be overcome by assuming, wlog, that $B \geq C$. (Otherwise, we do the argument based on $\angle C$.) Note that, for $O$ inside $\triangle ABC$, $$\angle BAT \geq \angle BAH \iff \frac{1}{2}A \geq 90^\circ - B \iff A \geq(A+B+C)-2B \iff B \geq C$$
– Blue
Commented Jan 10, 2014 at 20:42
• I agree, but perhaps you should add the wlog to your proof. Commented Jan 11, 2014 at 0:00

Going diameter $AD$ and analyzed angles $BAH$ and $CAD$.