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 $OAH$.
[Math] Property of angle bisectors in a triangle
geometrytriangles
Related Solutions
If 1 and 2 are the same, then the angle bisect or theorem tells us that the third sides are equal, hence the triangle is isosceles.
Here is a possible picture for the already solved problem:
Well, how could you fill in the details without a faithful picture?! (And what is the "reflection of $N$ across $S$" by construction, the point $N$ is reflected with respect to the center $S$ or conversely?!)
Best Answer
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 $$\angle HAT = \angle BAT + \angle BAH= \angle CAT + \angle OAD = \angle OAT.$$