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i do need another pair of eyes to look at this stupid question:

Take a regular pentagon and draw diagonals from each vertex/point. By doing that, you create another pentagon in the middle of the old one. Regarding the n-Pentagon with sides $s_n$ and diagonals $d_n$, show the following formula:

$d_{n+1} = d_n - s_n$

$s_{n+1} = d_n - 2d_{n+1} = 2s_n - d_n$

As mentioned, im not to interested in any kind of full solutions, since this is pretty basic stuff, i just in the need for someone to make sense of this question and explain to me, what i have to do here.

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    It looks like the construction in the first paragraph is repeated recursively. I.e., you start with a regular pentagon, construct a pentagon inside of it, construct another inside of that one, and so on. The formulas they want you to prove are relations between the lengths of the sides and diagonals of successive pentagons in this construction. – 2017-01-15

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Given a regular pentagon $ABCDE$, draw diagonals $AC$ and $BE$ which intersect at $F$. $BCDF$ is a parallelogram because each diagonal is parallel to the side it does not meet at the vertices, thus $AF=BF=$(diagonal minus side).

Next draw all five diagonals of the pentagon. Let $F$ be the intersection of $AC$ and $BE$ as above; $G$ be the intersection of $AD$ and $BE$; $H$ be the intersection of $BD$ and $CE$. You should be able to prove triangles $AFG$ and $HFG$ congruent by $ASA$, thus the diagonals of the inner pentagon equal (diagonal minus side) of the outer one.

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    Nice, I added a diagram. By symmetry $\triangle AFG$ and $\triangle HFG$ are both isosceles and because $F,G$ are on the diagonal of parallelogram $ABHE$ the altitudes match $\implies$ congruent. – 2017-01-15
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    Nice diagram @Joffan. – 2017-01-15