GS. 2 In the figure, $A B C D$ is a quadrilateral, where the interior angles $\angle A, \angle B$ and $\angle D$ are all equal to $45^{\circ}$. When produced, $B C$ is perpendicular to $A D$. If $A C=10$ and $B D=b$, find the value of $b$.
reflex $\angle B C D=360^{\circ}-45^{\circ}-45^{\circ}-45^{\circ}=225^{\circ}(\angle \mathrm{s}$ sum of polygon $)$
$\angle B C D=360^{\circ}-225^{\circ}=135^{\circ}(\angle \mathrm{s}$ at a point $)$
Produce $B C$ to meet $A D$ at $E, \angle A E B=90^{\circ}$ (given)
$$
\angle B A E=45^{\circ}=\angle A B E \text { (given) }
$$
$\triangle A B E$ and $\triangle C D E$ are right angled isosceles triangles
Let $A E=x, D E=y$, then $B E=x, C E=y, B C=x-y$
In $\triangle A C E, x^{2}+y^{2}=10^{2} \ldots$ (1) (Pythagoras' theorem)
$C D=\sqrt{y^{2}+y^{2}}=\sqrt{2} y \quad$ (Pythagoras' theorem)
Apply cosine rule on $\triangle B C D$
$$
\begin{array}{l}
B D^{2}=(x-y)^{2}+2 y^{2}-2(x-y) \sqrt{2} y \cos 135^{\circ} \\
B D^{2}=x^{2}-2 x y+y^{2}+2 y^{2}+2(x-y) y=x^{2}+y^{2}=10^{2} \\
\Rightarrow B D=b=10
\end{array}
$$