We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ . 
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .

What I Tried: Here is a picture :-

Let $AC = CD = x$. As $\angle ACD = 90^\circ$, we have $AD$ the diameter of the circle, so $AD = 2r$.
From here, using Pythagorean Theorem :- $$2x^2 = 4r^2$$
$$\rightarrow x = r\sqrt{2}$$
We have the green angles to be $45^\circ$, now as $ADCB$ is cyclic, $\angle ABC = 135^\circ$ and each of the brown angles are $22.5^\circ$. So we can use :-
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\rightarrow a \div \frac{\sqrt{2 - \sqrt{2}}}{2} = r\sqrt{2} * \frac{2}{\sqrt{2}}$$
$$\rightarrow \frac{2a}{\sqrt{2 - \sqrt{2}}} = 2r$$
$$BC = a = r\sqrt{2 - \sqrt{2}}$$
However, the answer to my question is given as $r\sqrt{\sqrt{2}}$ , so where did I go wrong?

The given answer is larger than $r$, which clearly isn't true, as mirroring $B$ across $AC$ shows $|AB|<|AO|$, where $O$ is the center of the circle.
