6. A line cuts off triangle $A K N$ from a regular hexagon $A B C D E F$ such that $A K+A N=A B$. Find the sum of the angles under which segment $K N$ is seen from the vertices of the hexagon ( $\angle K A N+\angle K B N+\angle K C N+\angle K D N+\angle K E N+$ $+\angle K F N$).

## 9 t h  g r a d e
6. Let's assume that $N$ lies on $A B$, and $K$ lies on $A F$ (Fig. 60). Note that $F K = A N$. We choose point $P$ on $B C$, point $R$ on $C D$, point $S$ on $D E$, and point $T$ on $E F$ such that the equalities $F K = A N = B P = C R = D S = E T$ hold. Then $\angle K B N = \angle T A K$, $\angle K C N = \angle S A T$, $\angle K D N = \angle R A S$, $\angle K E N = \angle P A R$, $\angle K F N = \angle N A P$, from which

$$
\begin{gathered}
\angle K A N + \angle K B N + \angle K C N + \angle K D N + \\
+ \angle K E N + \angle K F N = \\
= \angle K A N + \angle T A K + \angle S A T + \angle R A S + \\
+ \angle P A R + \angle N A P = \\
= \angle K A N + \angle K A N = 120^{\circ} + 120^{\circ} = 240^{\circ}.
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-150.jpg?height=371&width=331&top_left_y=1177&top_left_x=916)

Fig. 60

9th grade