# Regular Pentagon sides in terms of interior pentagon and segments connecting vertices

Diagonals of a regular pentagon $P$ form smaller interior pentagon $Q$. Let $d$ be the distance from any vertex of $P$ to the nearest vertex of $Q$. $a$ is the length of a side of $P$ and $b$ is the length of a side of $Q$. Prove that $$d=(ab)^{1/2} = \sqrt{ab}$$ For some reason, my brain is glitching on this seemingly easy geometry problem. I know that by the property of similar triangles, $$\frac{d}{b}=\frac{2d+b}{a}$$ Additionally, the interior angles are $108^\circ$, which when trisected to create smaller right triangle will be $\dfrac{108}{3}^\circ$. This would help in finding $d$, $a$, and $b$ in terms of trigonometry. However, this problem seems like it can be solved with similar triangles OR Pythagoras Theorem. Thank you in advance.

You can pick out another pair of similar triangles which gives $$\frac{a}{d}=\frac{2d+b}{a}.$$ Thus $$\frac{a}{d} = \frac{d}{b}.$$ The pair is: triangle formed by two adjacent sides and a diagonal of the big pentagon; and triangle formed by two edges joining a vertex of the small pentagon and two nearest vertices of the big pentagon and one side of the big pentagon.