IRL ABooklet

We connect $C$ to $A$ and $E$. Because triangles $ABC$ and $CDE$ are isosceles with an angle of $120^\circ$, we have $\angle BAC = \angle BCA = \angle DCE = \angle DEC = 30^\circ$.

Hence, $\angle ACE = 40^\circ$. Let $x = |AE|$ and $d = |CA| = |CE|$. The Cosine rule for $\triangle CDE$ gives $d^2 = 2 - 2\cos(120^\circ) = 3$. The Cosine Rule for $\triangle ACE$ then says $x^2 = 2d^2 - 2d^2 \cos(40^\circ) = 6(1 - \cos(40^\circ))$, and so $x > 1$ iff $\cos(40^\circ) < \frac{5}{6}$. To show this, we consider a regular pentagon with side length 1 and draw two diagonals that start at the same vertex. Because the internal angles of a regular pentagon have measure $108^\circ$, the angles are as indicated in the diagram below. Our aim is to prove $\cos(36^\circ) < 5/6$.

Letting $y = |CE| = |CA|$ and considering the perpendicular bisector of $EA$ in triangle $ACE$, we obtain $\cos(72^\circ) = \cos(\angle AEC) = \frac{1}{2y}$. The Cosine Rule for triangle $ACE$ gives $1 = 2y^2 - 2y^2 \cos(36^\circ)$, hence $\cos(36^\circ) = 1 - \frac{1}{2y^2}$. On the other hand, the Cosine Rule for triangle $ABC$ tells us $y^2 = 2 - 2\cos(108^\circ)$. Using $\cos(72^\circ) = -\cos(108^\circ)$, this implies $y^2 = 2 + 2\cos(72^\circ) = 2 + \frac{1}{y}$. Because $y$ is opposite the largest angle in triangle $ABC$, it must be its largest side, i.e. $y > 1$. Therefore, $y^2 = 2 + \frac{1}{y} < 3$. This then implies
$$\cos(36^\circ) = 1 - \frac{1}{2y^2} < 1 - \frac{1}{6} = \frac{5}{6}.$$
Finally, we obtain $\cos(40^\circ) < \cos(36^\circ) < \frac{5}{6}$ which shows that $|EA| = x > 1$ in the originally given pentagon.

Remark. From triangle $ABC$ we obtain $1 = 1 + y^2 - 2y \cos(36^\circ)$ (Cosine Rule) and so $\cos(36^\circ) = \frac{y}{2}$. Using our previous value, we get $\frac{y}{2} = 1 - \frac{1}{2y^2}$, hence, after multiplying by $2y$, $y^2 = 2y - \frac{1}{y}$. Comparing this with an earlier calculation gives $2y - \frac{1}{y} = 2 + \frac{1}{y}$, i.e. $y^2 - y - 1 = 0$. The positive root of this quadratic is $y = \frac{1+\sqrt{5}}{2}$, the golden ratio. This gives us the precise value
$$\cos(36^\circ) = \frac{y}{2} = \frac{1 + \sqrt{5}}{4}$$
which can easily be shown to be smaller than $5/6$.