It's All Geometry!

The area of $\Delta DBE$ is $\frac12 DE \cdot AB$ .

Since $DE=x+1$ , this means that $AB=2x-2$ .

From $\Delta ABC$ , $\tan \alpha =\frac{2x-2}{3x+3}$

From triangles $\Delta ABD$ and $\Delta ABE$ , $\alpha=\tan^{-1} \frac{2x+1}{2x-2}-\tan^{-1} \frac{x}{2x-2}$

Using the tangent addition formula, and after a bit of paperwork, this becomes $\tan\alpha=\frac{2x^2-2}{6x^2-7x+4}$

Combining these results and tidying, we get the cubic $(x - 1) \left(3 x^2 - 13 x + 1\right)=0$

The only root that makes sense in the context of the problem is $x=\frac{13+\sqrt{157}}{6} \approx \boxed{4.255}$

$m = 180 - (\alpha + \beta) = 180 - \lambda \implies \lambda = \alpha + \beta$

$\implies$ $\tan(\lambda) = \dfrac{h}{x} = \tan(\alpha + \beta)$ , where $\tan(\beta) = \dfrac{h}{2x + 1}$ and $\tan(\alpha) = \dfrac{h}{3x + 3}$

$\implies$ $\dfrac{h}{x} = \tan(\alpha + \beta) = \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} =$ $\dfrac{\dfrac{h}{3x + 3} + \dfrac{h}{2x + 1}}{1 - \dfrac{h^2}{(3x + 3)(2x + 1)}} = (\dfrac{5x + 4}{6x^2 + 9x + 3 - h^2})h$

$\implies \dfrac{1}{x} = \dfrac{5x + 4}{6x^2 + 9x + 3 - h^2} \implies 6x^2 + 9x + 3 - h^2 = 5x^2 + 4x$

$\implies h^2 = x^2 + 5x + 3 \implies h = \sqrt{x^2 + 5x + 3}$

$\implies A_{\triangle{DBE}} = \dfrac{1}{2}(x + 1)(\sqrt{x^2 + 5x + 3}) = x^2 - 1 = (x - 1)(x + 1) \implies$

$\sqrt{x^2 + 5x + 3} = 2(x - 1) \implies x^2 + 5x + 3 = 4x^2 - 8x + 4 \implies$

$3x^2 - 13x + 1 = 0 \implies x = \dfrac{13 \pm \sqrt{157}}{6}$ and $x > 1 \implies x = \dfrac{13 + \sqrt{157}}{6}$ $\approx \boxed{4.254994}$ .

Since $A_{\triangle DBA} = x^2 - 1$ , $\implies AB = 2(x-1)$ so that $A_{\triangle DBA} = \dfrac 12 AB \cdot DE = \dfrac 12 \cdot 2(x-1)(x+1) = x^2 -1$ . The $BD = \sqrt{AB^2+AD^2} = \sqrt{(2(x-1))^2 + x^2} = \sqrt{5x^2 - 8x + 4}$ . Note that $\triangle BDE$ and $\triangle BCD$ are similar, then we have:

$\begin{aligned} \frac {BD}{DE} & = \frac {CD}{BD} \\ \frac {\sqrt{5x^2-8x+4}}{x+1} & = \frac {2x+3}{\sqrt{5x^2-8x+4}} \\ 5x^2-8x+4 & = 2x^2 + 5x + 3 \\ 3x^2 - 13x + 1 & = 0 \\ \implies x & = \begin{cases} \dfrac {13+\sqrt{157}}6 \approx 4.254994014 \\ \dfrac {13-\sqrt{157}}6 \approx 0.078339319 \end{cases} \end{aligned}$

From $A_{\triangle DBE} = x^2 -1 \implies x > 1$ , therefore $x \approx \boxed{4.25}$ .