[The ratio in which the bisector divides the side, $]$ [

In triangle $A B C$, the median $A D$, the bisector $A E$, and the altitude $A F$ are drawn. The area of triangle $A E D$ is $\frac{1}{14}$ of the area of triangle $A B C$, and the area of triangle $A F D$ is $\frac{7}{50}$ of the area of triangle $A B C$. Find the angles of triangle $A B C$.

#
Let $B C=a, A C=b, A B=c(c<b)$. By the property of the angle bisector of a triangle $\frac{B E}{E C}=\frac{c}{b}$. Then

$$
B E=\frac{a c}{b+c}, D E=B D-B E=\frac{a}{2}-\frac{a c}{2+c}=\frac{a(b-c)}{2(b+c)}
$$

From the problem statement, it follows that $\frac{\underline{D E}}{B C}=\frac{1}{14}$, or

$$
\frac{\frac{a(b-c)}{2(b+c)}}{a}=\frac{b-c}{2(b+c)}=\frac{1}{14}
$$

from which $b=\frac{4}{3} c$. Further, we have:

$$
\begin{aligned}
& F D=\frac{7}{50} B C=\frac{7}{50} a, B F=B D-F D=\frac{\frac{2}{2}-50}{-50} a=\frac{9}{25} a \text {, }
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-22.jpg?height=88&width=500&top_left_y=2683&top_left_x=779)

$$
b 2-c 2=\left(\frac{16}{25 a} a\right) 2-\left(\frac{9}{25} a\right) 2, \frac{16}{9} c 2-c 2=\left(\frac{16}{25} a-\frac{9}{25} a\right)\left(\frac{16}{25 a}+\frac{9}{25} a\right),
$$

$$
\frac{7}{9} c 2=\frac{7}{25} a 2, a=\frac{5}{3} c
$$

Thus, the sides of triangle $A B C$ are proportional to the numbers 3, 4, 5. Therefore, triangle $A B C$ is a right triangle:

$$
\angle A=90^{\circ}, \angle B=\arccos \frac{3}{5}, \angle C=\arccos \frac{4}{5}
$$

## Answer

$\angle A=90^{\circ}, \angle B=\arccos \frac{3}{5}, \angle C=\arccos \frac{4}{5}$