2. Inside the square $A B C D$, there is a point $E$ such that the triangle $C D E$ is isosceles with an angle of $150^{\circ}$ at vertex $E$. Determine the angles of triangle $A B E$.

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Solution. Let $E_{1}$ be an internal point of the square $A B C D$ such that $A B E_{1}$ is an equilateral triangle, as shown in the diagram on the right. Then $C D E_{1}$ and $C E_{1} B$ are isosceles triangles (with bases $C D$ and $C E_{1}$, respectively), and we have

$$
\begin{aligned}
\measuredangle C E_{1} D & =180^{\circ}-2 \measuredangle E_{1} C D \\
& =180^{\circ}-2\left(90^{\circ}-\measuredangle E_{1} C B\right) \\
& =2 \measuredangle E_{1} C B=180^{\circ}-\measuredangle E_{1} B C \\
& =180^{\circ}-\left(90^{\circ}-60^{\circ}\right)=150^{\circ}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_6810ce1ac1ae234e68eeg-1.jpg?height=463&width=373&top_left_y=1148&top_left_x=1092)

Therefore, $E=E_{1}$, and all angles of triangle $A B E$ are equal to $60^{\circ}$.