How to prove angle bisector $\angle A$, is also angle bisector between height $AH$ and diameter $AM$ in circumscribed circle of the triangle

Question

How to prove that in $\triangle ABC$, angle bisector $\angle A$, is also angle bisector between height $AH$ and diameter $AM$ in circumscribed circle of the triangle.

Answer 1

Let $K$ be a midpoint of the arc $BC$ and $\gamma>\beta$.

If $\gamma=\beta$ then the lines $AM$, $AK$ and $AH$ are the same line and your statement is wrong.

We have: $\measuredangle AMK=\frac{\alpha}{2}+\beta$ and since $\measuredangle MKA=90^{\circ}$, we obtain $\measuredangle MAK=90^{\circ}-\frac{\alpha}{2}-\beta$.

In another hand, $\measuredangle MAH=\measuredangle BAH-\measuredangle BAM=90^{\circ}-\beta-\left(90^{\circ}-\gamma\right)=\gamma-\beta$.

Thus, we need to prove that $$2\left(90^{\circ}-\frac{\alpha}{2}-\beta\right)=\gamma-\beta$$ or $$\alpha+\beta+\gamma=180^{\circ}.$$ Done!

Answer 2

Hint:

$\angle CBM=\angle CAM$