Expressing vectors in an octagon I'm having trouble with this question in my course.
I am to consider a regular octagon with vertices A, B, C, D, E, F, G and H in counter clockwise order. The vectors $\overrightarrow{AC}$ and $\overrightarrow{AD}$ make up a base for the plane. Then I am supposed to express the vectors $\overrightarrow{AB}$, $\overrightarrow{AE}$, $\overrightarrow{AF}$ and $\overrightarrow{AG}$ in this base.
I've drawn the octagon with sides of unit length and divided it into eight isosceles triangles with angles $\frac{\pi}{4}$ and $\frac{3\pi}{8}$ (two of these). Through some calculations I came up with that the distance connecting two directly opposite points vertically is equal to $\tan\left(\frac{3\pi}{8}\right)$ and that the diagonals (for example A to E) are equal to $\sqrt{4+2\sqrt{2}}$. I'm thinking of finding some kind of ratio between the base vectors and the vertical line or diagonals that might help me solve this, but I am not sure how to proceed.
Need some good guidance.
Many thanks,
 A: Hint:
$\vec{HG}=\vec{CD} = \vec{AD} - \vec{AC}$
$\vec{CD} || \vec{AF} \Rightarrow \vec{AF} = k\vec{CD}$
$\vec{AD} || \vec{BC} \Rightarrow \vec{AD} = k'\vec{BC}$
$\vec{AB} = \vec{AC} - \vec{BC} $
$\vec{AE} = \vec{AF} + \vec{AB} $
$\vec{AE} = \vec{AC} + \vec{AG} $
A: I'll show you how you can do it for $AB$ and then you can take it from there:
$AB = AC + CB = AC - BC$
$BC = (\sqrt{2}-1)AD$
$\implies AB = AC - (\sqrt{2}-1)AD$
You get the factor $\sqrt{2} - 1$ by noticing that if each side of the octagon has length 1, then $|AD| = 1 + \sqrt{2}$. Since $BC$ has the same direction as $AD$, and has length 1, it can be expressed as:
$ \frac{1}{1 + \sqrt{2}}AD = (\sqrt{2}-1)AD$
A: Here is the complete solution:
$AB = AC + CB = AC - BC$
$BC = (\sqrt{2}-1)AD$
$\implies AB = AC - (\sqrt{2}-1)AD$
$AE = AB + AF$
$CD = (\sqrt{2}-1)AF$
$CD = CA + AD = -AC + AD$
$\implies AF = (\sqrt{2}+1)(AD - AC)$
$\implies AE = AB + (\sqrt{2}+1)(AD - AC)$
$\implies AE = AC - (\sqrt{2}-1)AD + (\sqrt{2}+1)(AD - AC)$
$\iff AE = -\sqrt{2}AC + 2AD$
$AF = AE - AB =  -\sqrt{2}AC + 2AD - AC + (\sqrt{2}-1)AD$
$\iff AF = (-\sqrt{2}-1)AC + (\sqrt{2}+1)AD$
$AE = AC + AG$ 
$\iff AG = AE - AC = (-\sqrt{2}-1)AC + 2AD$
