1. In rectangle $A B D F$, on sides $B D=2$ and $D F=3$, points $C$ and $E$ are chosen respectively, such that triangle $A F E$ is equal to triangle $E D C$. Then, from rectangle $A B D F$, triangles $A B C$, $C D E$, and $A F E$ are cut off. Find the angles of the remaining triangle.
Answer: $45^{\circ}, 45^{\circ}, 90^{\circ}$. Solution. Since triangles $A F E$ and $E D C$ are equal, then $\angle F A E=\angle D E C$, $\angle F E A=\angle D C E, A E=E C$. At the same time, since $\angle F E A+\angle F A E=90^{\circ}$, then $\angle F E A+\angle D E C=90^{\circ}$. Therefore, $\angle A E C=90^{\circ}$ and triangle $A E C$ is isosceles. Therefore, its angles are: $45^{\circ}, 45^{\circ}, 90^{\circ}$.

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