Six, as shown in Figure 4, in the cyclic quadrilateral $\mathrm{ABCD}$, $\angle \mathrm{B}$ $=\angle \mathrm{D}=90^{\circ}, \mathrm{AB}=2 \mathrm{~cm}, \mathrm{CD}=1 \mathrm{~cm}, \angle \mathrm{A}=45^{\circ}$. Find the lengths of $B C$ and $A D$. (10 points)
Six, Solution: As shown in Figure 8, connect $\mathrm{AC}$, let $\mathrm{BC}$ be $\mathrm{xcm}$, $A D$ be $\mathrm{ycm}$.
$$
\begin{array}{l}
\text { In Rt } \triangle \mathrm{ABC} \text { and } \mathrm{R}_{\mathrm{t}} \\
\triangle \mathrm{ADC} \text {, } \\
\mathrm{x}^{2}+2^{2}=\mathrm{AC}^{2}, \\
\mathrm{y}^{2}+1^{2}=\mathrm{AC}^{2},
\end{array}
$$

Thus, $x^{2}+4=y^{2}+1$,
$$
x^{2}-y^{2}=-3 \text {. }
$$

For $B D$, in: $\triangle A B D$ and $\triangle \mathrm{CBD}$,
$$
\begin{aligned}
\mathrm{BD}^{2} & =\mathrm{AB}^{2}+\mathrm{AD}^{2}-2 \mathrm{AB} \cdot \mathrm{AD} \cos \mathrm{A} \\
& =4+\mathrm{y}^{2}-4 y \cos 45^{\circ}, \\
\mathrm{BD}^{2} & =\mathrm{CD}^{2}+\mathrm{BC}^{2}-2 \mathrm{CD} \cdot \mathrm{BC} \cos \mathrm{C} \\
& =1+\mathrm{x}^{2}-2 \mathrm{x} \cos 135^{\circ},
\end{aligned}
$$

Thus, $4+y^{2}-4 y \cos 45^{\circ}=1+x^{2}-2 x \cos 135^{\circ}$,
$$
\begin{array}{l}
x^{2}-y^{2}+\sqrt{2} x+2 \sqrt{2} y=3 . \\
\because x^{2}-y^{2}=-3, \quad \therefore 2 x+2 \sqrt{2} y=6 .
\end{array}
$$

Solving the system: $\left\{\begin{array}{l}\sqrt{2}+2 \sqrt{2} y=6, \\ x^{2}-y^{2}=-3 .\end{array}\right.$
From (1), we get $x=3 \sqrt{2}-2 y$.
Substituting (3) into (2), we get $(3 \sqrt{2}-2 y)^{2}-y^{2}=-3$,
$$
\begin{array}{l}
\mathrm{y}^{2}-4 \sqrt{2} y+7=0 \text {, } \\
\mathrm{y}=\frac{4 \sqrt{2} \pm 2}{2} \text {. } \\
y_{1}=2 \sqrt{2}+1, y_{2}=2 \sqrt{2}-1 \text {. } \\
\text { Substituting into (3), we get } x_{1}=-\sqrt{2}-2 \text {, (not valid) } \\
x_{2}=2-\sqrt{2} . \\
\therefore\left\{\begin{array}{l}
x=2-\sqrt{2} \\
y=2 \sqrt{2}-1
\end{array}\right. \\
\end{array}
$$

Answer: $\mathrm{BC}$ is $(2-\sqrt{2}) \mathrm{cm}$, and $\Lambda \mathrm{D}$ is $(2 \sqrt{2}-1) \mathrm{cm}$.