1. **State the problem:** We have a regular pentagon with side length 1 and a diagonal of length $x$. Using the symmetry and similar triangles inside the pentagon, we want to find a relationship between $x$ and 1, then find the exact length of the diagonal, and finally show the pentagon is constructible. 2. **Use similar triangles to find the ratio:** The problem states that from the symmetry of the pentagon, the triangles formed imply $$\frac{x}{1} = \frac{1}{x - 1}.$$ This comes from the fact that the diagonal $x$ and side 1 form two similar triangles where corresponding sides are proportional. 3. **Solve the equation for $x$:** Start with $$\frac{x}{1} = \frac{1}{x - 1}.$$ Cross-multiply: $$x(x - 1) = 1.$$ Expand: $$x^2 - x = 1.$$ Bring all terms to one side: $$x^2 - x - 1 = 0.$$ 4. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-1$, $c=-1$. Calculate the discriminant: $$\sqrt{(-1)^2 - 4 \times 1 \times (-1)} = \sqrt{1 + 4} = \sqrt{5}.$$ So, $$x = \frac{1 \pm \sqrt{5}}{2}.$$ Since $x$ is a length, it must be positive and greater than 1, so we take the positive root: $$x = \frac{1 + \sqrt{5}}{2}.$$ 5. **Conclude the diagonal length:** Each diagonal of the regular pentagon has length $$\boxed{\frac{1 + \sqrt{5}}{2}}.$$ This number is known as the golden ratio, often denoted by $\phi$. 6. **Show the pentagon is constructible:** A regular polygon is constructible with ruler and compass if the number of sides $n$ satisfies that $n$ is a product of a power of 2 and distinct Fermat primes. Since 5 is a Fermat prime, the regular pentagon is constructible. Also, the diagonal length involving $\sqrt{5}$ shows that the coordinates of the vertices can be expressed using square roots, which are constructible numbers. Hence, the regular pentagon is constructible.