Equal angles within a right triangle

Let $\angle ADE=\angle BEC = \theta$ . Then $\tan \theta = \tan \angle ADE = \dfrac x6 \implies \sin \theta = \dfrac x{\sqrt{x+6^2}}$ . Since $\triangle BEC$ and $\triangle EAD$ has the same height $x$ , the ratio of their area is given by:

$\begin{aligned} \frac {[BEC]}{[EAD]} & = \frac {\overline{BC}}{\overline{AD}} = \frac 26 = \frac 13 \\ 3[BEC] & = [EAD] \\ 3 \times \frac 12 \times \overline{BE} \times \overline{CE} \times \sin \theta & = \frac 12 \times \overline{AD} \times \overline{AE} \\ 3 \times \sqrt{x^2+1^2} \times \sqrt{x^2+3^2} \times \frac x{\sqrt{x^2+6^2}} & = 6x \\ \sqrt{(x^2+1)(x^2+9)} & = 2 \sqrt{x^2+36} \\ x^4 + 10x^2 + 9 & = 4x^2 + 144 \\ x^4 + 6x^2 - 135 & = 0 \\ (x^2-9)(x^2+15) & = 0 \\ \implies x^2 & = 9 & \small \blue{\text{Since }x^2 > 0} \\ x & = \boxed 3 \end{aligned}$

Let $\angle {AEB}=α$ and $\angle {BEC}=\angle {ADE}=β$ . Then $\tan α=\dfrac{1}{x}$ , $\tan (α+β)=\dfrac{3}{x}$ and $\tan β=\dfrac{x}{6}$ . Using the identity $\tan {(α+β)}= \dfrac{\tan α+\tan β}{1-\tan {α}\tan {β}}$ , we get $x=\boxed 3$