Given a regular hexagon $A B C D E F$. It is known that $\overrightarrow{A B}=\vec{a}, \overrightarrow{A F}=\vec{b}$. Find the vectors $\overrightarrow{A D}, \overrightarrow{B D}$, $\overrightarrow{F D}$ and $\overrightarrow{B M}$, where $M-$ is the midpoint of side $E F$.

#
For any three points $X, Y$ and $Z$, the following equality holds:

$$
\overrightarrow{X Y}=\overrightarrow{X Z}+\overrightarrow{Z Y}
$$

## Solution

$$
\begin{gathered}
\overrightarrow{A D}=\overrightarrow{A B}+\overrightarrow{B E}+\overrightarrow{E D}=\vec{a}+2 \vec{b}+\vec{a}=2 \vec{a}+2 \vec{b} \\
\overrightarrow{B D}=\overrightarrow{B E}+\overrightarrow{E D}=2 \vec{b}+\vec{a} ; \overrightarrow{F D}=\overrightarrow{F C}+\overrightarrow{C D}=2 \vec{a}+\vec{b}
\end{gathered}
$$

Let $O$ be the center of the given hexagon, and $N$ be the midpoint of $O F$. Then

$$
\overrightarrow{B M}=\overrightarrow{B O}+\overrightarrow{O N}+\overrightarrow{N M}=\vec{b}-\frac{1}{2} \vec{a}+\frac{1}{2} \vec{b}=-\frac{1}{2} \vec{a}+\frac{3}{2} \vec{b}
$$

## Answer

$\overrightarrow{A D}=2 \vec{a}+2 \vec{b}, \overrightarrow{B D}=\vec{a}+2 \vec{b}, \overrightarrow{F D}=2 \vec{a}+\vec{b}, \overrightarrow{B M}=-\frac{1}{2} \vec{a}+\frac{3}{2} \vec{b}$.