Solution:
First note that triangle $ABE$ is isosceles because $AB = BE$.

Let $x = \angle DEC$. Angle chasing gives:
$$x = \angle DEC = \angle BEA = \angle EAB = \angle ACD.$$
Therefore triangle $CDE$ is isosceles. Hence $DE = DC$. Since we are also given $AC = DE$ this implies $AC = DC$. Therefore $\triangle ACD$ is isosceles. Since $x = \angle ACD$ this gives us
$$\angle CDA = \angle DAC = 90^\circ - \frac{x}{2}$$
Now let $\lambda$ be the angle bisector of $\angle BAC$. Since $x = \angle BAC$ we know that $\lambda$ makes an angle of $\frac{1}{2} \angle BAC = \frac{x}{2}$ with line $AC$. Therefore the angle between $\lambda$ and $AD$ is
$$\frac{x}{2} + \angle DAC = \frac{x}{2} + \left(90^\circ - \frac{x}{2}\right) = 90^\circ$$
as required.