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: $$\begin{array}{rl}\frac{[△ABE]}{[△ACE]}& =\frac{EB}{EC}=\frac{AB}{AC}\\ \frac{[△ABE]}{[△ACE]}+1& =\frac{AB}{AC}+1\\ \frac{[△ABE]+[△ACE]}{[△ACE]}& =\frac{AB+AC}{AC}\\ \frac{[△ABC]}{[△ACE]}& =\frac{6+10}{10}=\frac{8}{5}.\end{array}$$ Now, we see that $[△ABC]=\frac{1}{2}\cdot 6\cdot 10=30,$ so $[△ACE]=\frac{5}{8}\cdot [△ABC]=\frac{5}{8}\cdot 30=\boxed{\frac{75}{4}}.$