Expressing vectors in an octagon

Question

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,

Answer 1

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} $

Answer 2

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$