Here is my take on this question.
Let vector
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD})
and
Exploit the fact that for a regular octagon suppose
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{AB}=\vec{a})
then theoretically
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{GH}=\vec{a})
and for a regular octagon we see that
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{AB}=\overrightarrow{BC}=\overrightarrow{CD}=\overrightarrow{HG}=\overrightarrow{GF}=\overrightarrow{FE})
because these vectors are going in the same direction.
As a result of out first statement, we said that
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD})
and
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{HE}=\overrightarrow{HG}+\overrightarrow{GF}+\overrightarrow{FE})
.
Using that we can say
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{AD}=\overrightarrow{HE})
.