$\triangle A B C$ is a right angled isosceles triangle.
$$
\begin{array}{l}
\angle B A C=45^{\circ}(\angle \mathrm{s} \text { sum of } \triangle \text {, base } \angle \mathrm{s} \text { isos. } \triangle \text { ) } \\
\left.\angle A C D=45^{\circ} \text { (alt. } \angle \mathrm{s}, A B / / D C\right) \\
\angle B C D=135^{\circ}
\end{array}
$$

Apply sine law on $\triangle B C D$,
$$
\begin{array}{l}
\frac{B D}{\sin 135^{\circ}}=\frac{B C}{\sin D} \\
A B \sqrt{2}=\frac{A B \sin 45^{\circ}}{\sin D}, \text { given that } A B=B D \\
\sin D=\frac{1}{2} ; D=30^{\circ} \\
\angle C B D=180^{\circ}-135^{\circ}-30^{\circ}=15^{\circ}(\angle \mathrm{s} \text { sum of } \triangle B C D) \\
b=15
\end{array}
$$
SG. 2 In figure $1, A B$ is parallel to $D C, \angle A C B$ is a right angle, $A C=C B$ and $A B=B D$. If $\angle C B D=b^{\circ}$, find the value of $b$.