Segments Within Rectangles

As alternate interior angles to parallel lines $AB$ and $DC$ , $\angle AED \cong \angle CDE$ , and so $\angle CED \cong \angle AED \cong \angle CDE$ which means $\triangle CED$ is an isosceles triangle with $CD \cong CE$ .

Since we are given that $CD = 2 \cdot BC$ , and since $CD \cong CE$ , we have $CE = 2 \cdot BC$ . Using right triangle $\triangle BEC$ , $\sin \angle BEC = \frac{BC}{CE}$ or $\sin \angle BEC = \frac{1}{2}$ , which means $\angle BEC = 30°$ .

Finally, as a straight angle, $\angle AEB = 180°$ , and since $\angle AEB = \angle AED + \angle CED + \angle BEC$ and $\angle AED \cong \angle CED$ , we have $180° = \angle AED + \angle AED + 30°$ , which means $\angle AED = \boxed{75°}$ .

Let $\overline{AD}=\overline{BC}=1$ , then $\overline{AB} = \overline{CD}=2$ , and $\angle AED = \angle DEC = \theta$ . Since $\overline{AB}$ is parallel to $\overline{CD}$ , $\angle EDC = \angle AED = \theta$ and $\triangle CDE$ is an isosceles triangle. Therefore, $\overline{CE} = \overline{CD} = 2$ .

We note that $\overline{ED} = \dfrac {\overline{AD}}{\sin \angle AED} = \dfrac 1{\sin \theta}$ . Also $\overline{ED} = 2\overline{CE} \cos \angle DEC = 2\times 2\cos \theta$ . Therefore, we have:

$\begin{aligned} 4 \cos \theta & = \frac 1{\sin \theta} \\ 2\sin \theta \cos \theta & = \frac 12 \\ \sin (2\theta) & = \frac 12 \\ \implies 2 \theta & = 150^\circ & \small \color{#3D99F6} \text{Note that }2 \theta > 30^\circ \\ \theta & = \boxed{75}^\circ \end{aligned}$

Clue: Compute that triangle EBC is 30 60 90 triangle

Tha angle $EDC$ is congruent to angle $AED$ couse segments $AB$ and $DC$ are parallel and $DE$ in a trasversal. Hence angles $DEC$ and $EDC$ are congruent and the triangle $ECD$ is isosceles. Consider the right triangle $CBE$ (right in $B$ ). It is half an equilateral triangle couse its hypotenuse $EC$ is double the side $BC$ . So angle $BCE$ is $60^{\circ}$ and angle $CEB$ is $30^\circ$ . So angle $AEC$ is $150^\circ$ , and angle $AED$ is $75^\circ$ .

Let AE = r. Let <AED = t. Computing angles, we find < BCE = 2t - 90. Then in triangle AED, tan(t) = 1/r. In triangle BCE, tan(2t - 90) = 2 -r, where we have assumed the dimensions of the rectangle are 1 and 2. So sin(t)/cos(t) = 1/r, and sin(2t - 90)/cos(2t - 90) = -cos(2t)/sin(2t) = {sin^(t) - cos^(t)/(2sin(t)cos(t) = 2 - r. Substituting cos(t) = r*sin(t), and simplifying, r^2 - 4r + 1 =0, With root r = 2 - sqrt(3), then tan(t) = 1/r gives t = 75. Ed Gray]

Suppose $|AD| = a, |DC| = 2a$ , and $\angle AED = \angle DEC = \theta$ . Then $\angle BEC = \pi - 2\theta$ . So with $|AE| = x$ we have

$\cot(\angle AED) = \cot(\theta) = \dfrac{x}{a}$ and $\cot(\angle BEC) = \cot(\pi - 2\theta) = \dfrac{2a - x}{a} = 2 - \dfrac{x}{a}$ , in which case

$\cot(\theta) + \cot(\pi - 2\theta) = 2 \Longrightarrow \cot(\theta) - \cot(2\theta) = 2 \Longrightarrow \cot(\theta) - \dfrac{1 - \tan^{2}(\theta)}{2\tan(\theta)} = 2$

$\Longrightarrow 2 - (1 - \tan^{2}(\theta)) = 4\tan(\theta) \Longrightarrow \tan^{2}(\theta) - 4\tan(\theta) + 1 = 0 \Longrightarrow \tan(\theta) = \dfrac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$ .

Now clearly $\theta \gt \pi /4 \Longrightarrow \tan(\theta) \gt 1$ , so we take the positive root, i.e., $\tan(\theta) = 2 + \sqrt{3}$ , from which we recognize that $\theta = \boxed{75^{\circ}}$ .

Note: $\tan(75^{\circ}) = \tan(45^{\circ} + 30^{\circ}) = \dfrac{\tan(45^{\circ}) + \tan(30^{\circ})}{1 - \tan(45^{\circ}) \tan(30^{\circ})} = \dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{\sqrt{3}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = \dfrac{(\sqrt{3} + 1)^{2}}{2} = \dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$ .