3. (K. Knop) On the side of a regular octagon, a square is constructed outward. In the octagon, two diagonals intersecting at point $B$ are drawn (see figure). Find the measure of angle $A B C$.

Note. A polygon is called regular if all its sides are equal and all its angles are equal.
Solution: First, note that the angle of a regular octagon is 6 * $180^{\circ} / 8=135^{\circ}$. Let's denote the vertices of the octagon as shown in the figure. Note that $K L D E$ is an isosceles trapezoid, so the angle $B E D$ is $45^{\circ}$, and the angle $F E B$ is $135^{\circ}-45^{\circ}=90^{\circ}$. Further, $H D$ is the axis of symmetry of the octagon, so the angle $H D E$ is $135^{\circ} / 2=67.5^{\circ}$. From this, it follows that in triangle $B D E$, angles $B$ and $D$ are equal, which means $E B=E D=E F=F C$. Triangles $B E F$ and $F C A$ are equal by two legs, so $B F=F A$. Next, the angle $F$ of the isosceles right triangle $B E F$ is $45^{\circ}$, so the angle $G F B$ is a right angle. The angle $A F B$ is the sum of angles $A F G$ and $G F B$, which is $135^{\circ}$. Now note that the sum of the equal angles $F B A$ and $F A B$ is $45^{\circ}$, so the angle $A B C$ is $45^{\circ} / 2=22.5^{\circ}$.

Answer: $22.5^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_825e0b293026a9348507g-2.jpg?height=511&width=372&top_left_y=30&top_left_x=862)