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

Question

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.

Answer 1

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.

Answer 2

similar triangles SUT and TSR along with the relation I found off the bat give the desired relation