jmc

Since $AB=BD,$ we see that $△ABD$ is an isosceles right triangle, therefore $\angle DAB=45^{\circ }.$ That means that $AD$, and consequently $AE,$ bisects $\angle CAB.$

Relating our areas to side lengths and applying the Angle Bisector Theorem, we have that: $$\begin{array}{rl}\frac{[△AEC]}{[△ABE]}& =\frac{CE}{EB}=\frac{CA}{AB}\\ \frac{[△AEC]}{[△ABE]}+1& =\frac{CA}{AB}+1\\ \frac{[△AEC]+[△ABE]}{[△ABE]}& =\frac{CA+AB}{AB}\\ \frac{[△ABC]}{[△ABE]}& =\frac{5+4}{4}=\frac{9}{4}.\end{array}$$ Now, we see that $[△ABC]=\frac{1}{2}\cdot 4\cdot 5=10,$ so $[△ABE]=\frac{4}{9}\cdot [△ABC]=\frac{4}{9}\cdot 10=\boxed{\frac{40}{9}}.$