What relationship exists between the angles of a triangle if the angle bisector from one vertex, the median from another vertex, and the altitude from the third vertex intersect at one point?
Let $A_{1}$ be any point on the side $BC$ of triangle $ABC$, and let the intersection of the line through $A_{1}$ parallel to $AC$ with $AB$ be $C_{1}$. The intersection of the lines $AA_{1}$ and $CC_{1}$ is $D$, and the points where the line through $D$ parallel to $AC$ intersects $BC$ and $AB$ are $A_{2}$ and $C_{2}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_ad7fd996fe7ba4f1c4d8g-1.jpg?height=518&width=892&top_left_y=305&top_left_x=606)

Then,

$$
A_{2} D=\frac{C D}{C C_{1}} A_{1} C_{1}, \quad C_{2} D=\frac{A D}{A A_{1}} A_{1} C_{1}
$$

and since $C D: D C_{1}=A D: D A_{1}$, it follows that $\frac{C D}{C C_{1}}=\frac{A D}{A A_{1}}$, thus $A_{2} D=C_{2} D$. Therefore, the midpoint $B_{1}$ of segment $AC$, $D$, and $B$ are collinear. Hence, if $D$ is defined as the intersection of $AA_{1}$ and $BB_{1}$, the line $CD$ can only intersect $AB$ at $C_{1}$. This can also be stated as follows: if the intersection of the lines $AA_{1}$ and $CC_{1}$ lies on the median $BB_{1}$, then

$$
B A_{1}: A_{1} C=B C_{1}: C_{1} A
$$

If $AA_{1}$ is the angle bisector and $CC_{1}$ is the altitude, then $B A_{1}: A_{1} C = AB: AC$ and $B C_{1}: C_{1} A = BC \cos \beta: AC \cos \alpha$, so (1) is equivalent to the condition

$$
\frac{AB}{AC} = \frac{BC \cos \beta}{AC \cos \alpha}
$$

which is equivalent to the condition

$$
\sin \gamma = \tan \alpha \cos \beta
$$