Let it intersect at some point $K$ then observe that $\angle AKB=\frac{\pi}{2}$ and write the relation of similarity in the $2$ triangles $CAB$ ,$DBA$. We get $\frac{AB}{AD}=\frac{BC}{AB}$ $AB=\sqrt{AD \cdot BC}$

Answer:$\sqrt {ab} $

Prooving that triangles are similar is also simple. $\angle CAD=\angle EBA$, $\angle CBD=\angle BAE$, since $\angle BAD=\angle BAE+ \angle CAD=90 ^\circ$, and $\angle BAE+ \angle ABE =90^\circ$, $\angle ABE+ \angle ADB = 90^\circ \Longrightarrow$ $\angle BAE=\angle ADB$ and triangles $ABD$ and $ABC$ are similar, so $\frac{AB}{a}=\frac{b}{AB}$ or $AB^2=ab$ and finally $AB=\sqrt {ab} $

I have only used Pythagoras. But the solution is big with the same answer √(ab).