G1.2 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 $b$.
$\triangle A B C$ is a right angled isosceles triangle.
$\angle B A C=45^{\circ}(\angle \mathrm{s}$ sum of $\triangle$, base $\angle \mathrm{s}$ isos. $\triangle$ )
$\angle A C D=45^{\circ}$ (alt. $\angle \mathrm{s}, A B / / D C$ )
$\angle B C D=135^{\circ}$
Apply sine law on $\triangle B C D$,
$\frac{B D}{\sin 135^{\circ}}=\frac{B C}{\sin D}$
$A B \sqrt{2}=\frac{A B \sin 45^{\circ}}{\sin D}$, given that $A B=B D$
$\sin D=\frac{1}{2} ; D=30^{\circ}$
$\angle C B D=180^{\circ}-135^{\circ}-30^{\circ}=15^{\circ}(\angle \mathrm{s}$ sum of $\triangle B C D)$
$b=15$