3. (15 points) The vertices of the broken line $A B C D E F G$ have coordinates $A(0 ; -5), B(3 ; 7), C(4 ; -6), D(-2 ; 6), E(6 ; 1), F(-3 ; 0), G(7 ; 6)$.

Find the sum of the angles with vertices at points $B, E, C, F, D$.
Answer: $135^{\circ}$.

Solution. The closed broken line $B C D E F B$ forms a five-pointed "star". The sum of the angles at the rays of this star is $180^{\circ}$.

We will prove that the sum of the angles at the rays of any five-pointed star $B C D E F B$ is $180^{\circ}$. Let $O$ be the point of intersection of the lines $B F$ and $E D$, and the angle between them is $B O D=\alpha$. Denote the angle at the ray by the same letter as the vertex of the ray.

$$
\angle E+\angle F=180^{\circ}-\alpha=\angle O B D+\angle O D B .
$$

Therefore,

$$
\begin{array}{r}
180^{\circ}=\angle C+\angle C B D+\angle C D B=\angle C+(\angle B+\angle O B D)+(\angle D+\angle O D B)= \\
\quad=\angle C+\angle B+\angle D+(\angle O B D+\angle O D B)=\angle C+\angle B+\angle D+\angle E+\angle F
\end{array}
$$

Another proof. Let the ray $\bar{a}$ coincide with the ray $В C$. Rotate the ray $\bar{a}$ until it coincides with the ray $B F$. The ray $\bar{a}$ will rotate by the angle $\angle B$. Then rotate the ray $\bar{a}$ (in its new position) until it coincides with the ray $E F$. The ray $\bar{a}$ will rotate by another angle $\angle F$, and from the start of the movement, by the angle $\angle B+\angle F$. Then rotate the ray $\bar{a}$ again, this time until it coincides with the ray $E D$. The ray $\bar{a}$ will rotate by another angle $\angle D$, and from the start of the movement, by the angle $\angle B+\angle F+\angle E$. After performing two more similar rotations, the ray $\bar{a}$ will coincide with the ray $C B$, i.e., it will have rotated by $180^{\circ}$ from the start of the movement and, at the same time, by the sum of the angles $\angle B+\angle F+\angle E+\angle D+\angle C$.

The point of intersection of the segments $A B$ and $F G$ is the point $K(2 ; 3)$.

We will prove this. Let $L(2 ; 0), M(7 ; 0)$. Then $\triangle K F L \sim \triangle G F M$, since their legs are proportional. Therefore, $\angle K F L=\angle G F M$, so $K \in F G$. Similarly, $K \in A B$.

Another way. The equation of the line $A B: y=4 x-5$. The equation of the line $F G: y=\frac{3}{5} x+\frac{9}{5}$. Solving the system of these two equations, we get the coordinates of the intersection point of these lines: $(2 ; 3)$.

Find the lengths of the sides of $\triangle B K G: B K=\sqrt{1^{2}+4^{2}}=\sqrt{17}, B G=$ $\sqrt{4^{2}+(-1)^{2}}=\sqrt{17}, G K=\sqrt{5^{2}+3^{2}}=\sqrt{34}$. The triangle $\triangle B K G$ is isosceles and, by the converse of the Pythagorean theorem, it is a right triangle.

Therefore, the angle $\angle B K G=45^{\circ}$. It is an exterior angle for $\triangle F K B$ and $\angle K F B+\angle K B F=45^{\circ}$. Since $\angle A B C=\angle F B C-\angle F B A$, $\angle E F G=\angle E F B-\angle G F B$, the required sum of the angles

$$
\begin{gathered}
\angle A B C+\angle E F G+\angle B C D+\angle C D E+\angle D E F= \\
=\angle F B C+\angle E F B-(\angle F B A+\angle G F B)+\angle B C D+\angle C D E+\angle D E F= \\
=180^{\circ}-45^{\circ}=135^{\circ}
\end{gathered}
$$

Remark. A similar solution can be obtained based on the fact that the sum of the angles at the rays of any seven-pointed "star" is also $180^{\circ}$ (the proof, in essence, does not differ from the second proof for a five-pointed star). Since the angle $\angle B K G=45^{\circ}$, the sum of the two angles of the seven-pointed star with vertices at points $A$ and $G$ is $45^{\circ}$.

Grading criteria. Full solution 15 points. If the coordinates of point $K$ are found from the drawing or guessed - minus 2 points. If the fact that $\angle B K G=45^{\circ}$ is found from the drawing or guessed - minus 3 points. If the problem is not solved, but it is shown that the sum of all angles at the rays of a five- or seven-pointed star is $180^{\circ}$, then 8 points.