explaining the resriction $b<a<2b$ in a triangle

Question

I saw in a book that if $ABC$ is an isosceles triangle $(AB=AC)$ and the triangle is tangent to a circle in points $D,C$ and $AC$ is intersecting the circle in point $E$; $AC=a$, $BC=b$

so it has the restriction: $b<a<2b$.

Can't see how they got that restriction and hope that someone could explain.

Thanks

EDIT: @shaurya gupta proved that $b<a$ but the hard part is to prove that $a<2b$

Answer 1

We have that $AB = AC = a$.
Clearly, $BD = BC$, because tangents drawn from common point to a circle are equal.
$\implies BD = BC = b$.
$AB = AD + BD $
$ \implies a = AD + b.$ So $b < a$