On the side $A B$ of triangle $A B C$, a circle is constructed with $A B$ as its diameter, intersecting sides $A C$ and $B C$ at points $D$ and $E$ respectively. The line $D E$ divides the area of triangle $A B C$ in half and forms an angle of $15^{\circ}$ with the line $A B$. Find the angles of triangle $A B C$.
Triangle $C D E$ is similar to triangle $C B A$ with a similarity coefficient of $|\cos \angle A C E|$.

## Solution

Let the angles at vertices $A, B$, and $C$ of triangle $A B C$ be denoted as $\alpha, \beta$, and $\gamma$ respectively. Since point $C$ lies outside the circle with diameter $A B$, the angle at vertex $C$ is acute, i.e., $\gamma<90^{\circ}$.

Points $A, D, E$, and $B$ lie on the same circle, therefore

$$
\angle C D E=180^{\circ}-\angle A D E=\angle A B C=\beta
$$

Similarly, we can prove that $\angle C E D=\alpha$.

Triangle $C D E$ is similar to triangle $C B A$, and the similarity coefficient is the square root of the ratio of the areas of these triangles, i.e., $k=\frac{1}{\sqrt{2}}$. On the other hand, since

$$
\angle A E C=\angle A E B=90^{\circ}
$$

then

$$
\frac{1}{\sqrt{2}}=k=\frac{C E}{A C}=\cos \angle A C E=\cos \gamma
$$

Therefore, $\gamma=45^{\circ}$.

Let the line $D E$ intersect the line $A B$ at point $K$. By the problem's condition, $\angle A K D=15^{\circ}$. At the same time, $\angle A B C$ is the external angle of triangle $B K$, so

$$
\angle A B C=\angle A K D+\angle B E K=\angle A K D+\angle C E D, \text { or } \beta=15^{\circ}+\alpha .
$$

Substituting $\beta$ into the equation

$$
\alpha+\beta=180^{\circ}-45^{\circ}=135^{\circ},
$$

we find that $\alpha=60^{\circ}$. Therefore, $\beta=75^{\circ}$.

## Answer

$60^{\circ}, 75^{\circ}, 45^{\circ}$.