\left.\begin{array}{l}{[\text { Sum of internal and external angles of a polygon }]} \\ {[\quad \text { Regular (equilateral) triangle }]}\end{array}\right]

In a convex hexagon $A B C D E F$, all internal angles at the vertices are equal. It is known that $A B=3, B C=4$, $C D=5$ and $E F=1$. Find the lengths of the sides $D E$ and $A F$.
Let the lines $A F$ and $B C$ intersect at point $K$, the lines $B C$ and $D E$ at point $L$, and the lines $A F$ and $D E$ at point $M$. Triangles $A K B, C L D, E M F$, and $K L M$ are equilateral.

## Solution

Let the lines $A F$ and $B C$ intersect at point $K$, the lines $B C$ and $D E$ at point $L$, and the lines $A F$ and $D E$ at point $M$. Since the sum of the interior angles of a convex hexagon is $180^{\circ}(6-2)=720^{\circ}$ and all these angles are equal, each of them is $120^{\circ}$. Therefore, triangles $A K B, C L D, E M F$, and $K L M$ are equilateral. Hence,

$$
\begin{gathered}
A K=K B=A B=3, C L=L D=C D=5, E M=M F=E F=1 \\
K M=M L=K L=K B+B C+C L=3+4+5=12 .
\end{gathered}
$$

Then

$$
D E=M L-M E-D L=12-1-5=6, A F=K M-A K-M F=12-3-1=8 .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-08.jpg?height=840&width=969&top_left_y=1172&top_left_x=545)

## Answer

6 and 8.