1. **State the problem:** We have a regular pentagon with side length 1 and a diagonal of length $x$. We want to use the symmetry and similar triangles to find a relationship between $x$ and 1, then find the length of the diagonal and show the pentagon is constructible. 2. **Use similar triangles:** The problem states the similarity implies $$\frac{x}{1} = \frac{1}{x - 1}$$ This comes from the ratio of corresponding sides in the two similar triangles formed by the diagonal and sides of the pentagon. 3. **Solve for $x$:** Cross-multiply: $$x(x - 1) = 1 \implies x^2 - x = 1$$ Rearranged: $$x^2 - x - 1 = 0$$ 4. **Solve quadratic equation:** Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-1$, $c=-1$: $$x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$$ Since $x$ is a length, it must be positive and greater than 1, so: $$x = \frac{1 + \sqrt{5}}{2}$$ 5. **Conclusion for part (ii):** Each diagonal of the regular pentagon has length $\frac{1 + \sqrt{5}}{2}$. 6. **Constructibility of the pentagon:** A regular pentagon is constructible with compass and straightedge because its side and diagonal lengths relate to the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$, which is a solution to a quadratic equation with rational coefficients. This means the coordinates of the vertices lie in a field extension of degree a power of 2, making the pentagon constructible. **Final answers:** (i) $\frac{x}{1} = \frac{1}{x - 1}$ (ii) $x = \frac{1 + \sqrt{5}}{2}$ (iii) The regular pentagon is constructible because its diagonal length is related to the golden ratio, a constructible number.