It's All Angles.

Expand the figure, to form square $ABFC$ . Denote the side length of the square by $s$ . $AD$ extended meets $CF$ at $M$ and $BE$ extended meets $AC$ at $N$ .

$\triangle ABD$ and $\triangle ABD$ are similar, thus, $\dfrac{CM}{AB}=\dfrac{DC}{BD}=\dfrac{1}{2}$ This gives $CM=\dfrac{AB}{2}=\dfrac{s}{2}$ , i.e. $M$ is the midpoint of $CF$ .

$\triangle ABN$ and $\triangle ACM$ are right-angled, with one pair of congruent legs ( $AB=AC$ ) and one pair of congruent acute angles ( $\angle ABN=\angle CAM$ because they are acute angles with perpendicular sides). Hence, $\triangle ABN\cong \triangle ACM$ thus, $AN=CM$ , i.e. . $N$ is the midpoint of $AC$ .

Since $\angle NCM=\angle NEM=90{}^\circ$ , quadrilateral $NCME$ is cyclic, thus, $\angle CEM=\angle CNM$

But, $NC=CM=\dfrac{s}{2}$ and $NCM=90{}^\circ$ , thus, $\triangle NCM$ is isosceles right triangle, hence, $\angle CNM=45{}^\circ$

Consequently, $\angle \lambda=\angle CNM=\boxed{45{}^\circ }$

Let $\overline{AB} = \overline{AC} = 1 \implies BC = \sqrt{2} \implies$ .

Using the above diagram $BC = \sqrt{2} = 3y \implies y = \dfrac{\sqrt{2}}{3} \implies \overline{BD} = \dfrac{2\sqrt{2}}{3}$ and using

$\triangle{BEC} \implies 45^{\circ} - \theta + 90^{\circ} + \alpha + 45^{\circ} - w = 180^{\circ} \implies \lambda = \theta + w$ .

Using the law of cosines on $\triangle{ABD}$ with included angle $90^{\circ} - \theta \implies$

$\overline{AD}^2 = 1 + \dfrac{8}{9} - \dfrac{4\sqrt{2}}{3}(\dfrac{1}{\sqrt{2}}) \implies \overline{AD} = \dfrac{\sqrt{5}}{3} \implies$

$\dfrac{8}{9} = 1 + \dfrac{5}{9} - \dfrac{2\sqrt{5}}{3}\sin(\theta) \implies \sin(\theta) = \dfrac{1}{\sqrt{5}} = \overline{AE} \implies$

$\overline{BE} = \cos(\theta) = \dfrac{2}{\sqrt{5}} \implies \overline{ED} = \dfrac{\sqrt{5}}{3} - \dfrac{1}{\sqrt{5}} = \dfrac{2}{3\sqrt{5}}$

Using the law of cosines on $\triangle{AEC}$ with included angle $\theta \implies$

$\overline{EC}^2 = 1 + \dfrac{1}{5} - \dfrac{4}{5} = \dfrac{2}{5} \implies \overline{EC} = \sqrt{\dfrac{2}{5}} \implies$

$\dfrac{1}{5} = 1 + \dfrac{2}{5} - \dfrac{2\sqrt{2}}{\sqrt{5}}\cos(w) \implies \cos(w) = \dfrac{3}{\sqrt{10}} \implies \sin(w) = \dfrac{1}{\sqrt{10}}$

$\implies \cos(\lambda) = \cos(\theta + w) = \dfrac{2}{\sqrt{5}}\dfrac{3}{\sqrt{10}} - \dfrac{1}{\sqrt{5}}\dfrac{1}{\sqrt{10}} = \sqrt{\dfrac{5}{10}} = \dfrac{1}{\sqrt{2}}$

$\implies \lambda = \boxed{45^{\circ}}$

Draw altitude $AF$ intersecting $BE$ at $G$ , and let $BF = AF = FC = 3$ so that $FD = 1$ and $DC = 2$ :

Then $\triangle BFG \sim \triangle BED$ by AA similarity, and $\triangle BED \sim \triangle AFD$ by AA similarity, and $\triangle AFD \sim \triangle AEG$ by AA similarity, so that all four triangles $\triangle BFG \sim \triangle BED \sim \triangle AFD \sim \triangle AEG$ .

Since $BF = AF$ and $\triangle BFG \sim \triangle AFD$ , $\triangle BFG \cong \triangle AFD$ , which means $GF = FD = 1$ . Then $AG = AF - GF = 3 - 1 = 2$ .

Let $x = GE$ . Since $\triangle AEG \sim \triangle AFD$ , $AE = EG \cdot \cfrac{AF}{FD} = x \cdot \cfrac{3}{1} = 3x$ .

Since $\triangle AEG \sim \triangle BED$ , $ED = BD \cdot \cfrac{GE}{AG} = 4 \cdot \cfrac{x}{2} = 2x$ .

Now copy and rotate $\triangle AFC$ and everything in it $90°$ clockwise about $F$ :

Then $\triangle CEE'$ is an isosceles right triangle, so $\angle CEE' = \lambda = \boxed{45°}$ .

Let $DC=1$ . Then $BD=2$ and $AB=AC = \dfrac 3{\sqrt 2}$ . Let $\angle BAD = \theta$ . Then be angle bisector theorem:

$\begin{aligned} \frac {AB \sin \angle BAD}{AC \sin \angle DAC} & = \frac {BD}{CD} & \small \blue {\text{Note that }AB=AC} \\ \frac {\sin \theta}{\sin (90^\circ - \theta)} & = \frac 21 \\ \frac {\sin \theta}{\cos \theta} & = 2 \\ \implies \tan \theta & = 2 \end{aligned}$

Then $\sin \theta = \dfrac 2{\sqrt 5}$ , $\cos \theta = \dfrac 1{\sqrt 5}$ , and $BE = AB \sin \theta = \dfrac 6{\sqrt{10}}$ . Let $EF$ be perpendicular to $BC$ . Then

$\begin{aligned} BF & = BE \cdot \cos (45^\circ - (90^\circ - \theta)) = BE \cdot \cos (\theta - 45^\circ) \\ & = \frac 6{\sqrt{10}} \left(\frac 1{\sqrt 2} \cdot \frac 1{\sqrt 5} + \frac 1{\sqrt 2}\cdot \frac 2{\sqrt 5} \right) = 1.8 \\ EF & = BE \cdot \sin (\theta - 45^\circ) = \frac 6{\sqrt{10}} \cdot \frac 1{\sqrt{10}} = 0.6 \end{aligned}$

And

$\begin{aligned} \lambda & = \angle FEC - \angle FED \\ \tan \lambda & = \tan (\angle FEC - \angle FED) \\ & = \frac {\tan \angle FEC-\tan \angle FED}{1-\tan \angle FEC \cdot \tan \angle FED} & \small \blue{\text{Note that }\tan \angle FEC = \frac {FC}{EF} = \frac {1.2}{0.6}=2} \\ & = \frac {2-\frac 13}{1+2 \cdot \frac 13} = 1 & \small \blue{\text{and }\tan \angle FED = \frac {FD}{EF} = \frac {0.2}{0.6} = \frac 13} \\ \implies \lambda & = \boxed{45}^\circ \end{aligned}$