解:作$OM \perp AB$于點(diǎn)$M,$連接$OA。$
因?yàn)?CD$為$\odot O$的直徑,$DE = 9\ \text{cm},$$CE = 3\ \text{cm},$所以直徑$CD=DE + CE=9 + 3=12\ \text{cm},$半徑$OA=\frac{1}{2}CD = 6\ \text{cm}。$
又因?yàn)?OD = OA=6\ \text{cm},$所以$OE=DE - OD=9 - 6=3\ \text{cm}。$
在直角$\triangle OEM$中,$\angle CEB = 45^\circ,$則$\angle OEM=45^\circ,$所以$OM = OE \cdot \sin45^\circ=3\times\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}\ \text{cm}。$
在直角$\triangle OAM$中,根據(jù)勾股定理得:$AM=\sqrt{OA^2 - OM^2}=\sqrt{6^2 - (\frac{3\sqrt{2}}{2})^2}=\sqrt{36 - \frac{9\times2}{4}}=\sqrt{36 - \frac{9}{2}}=\sqrt{\frac{72 - 9}{2}}=\sqrt{\frac{63}{2}}=\frac{3\sqrt{14}}{2}\ \text{cm}。$
因?yàn)?OM \perp AB,$由垂徑定理可知$AB = 2AM,$所以$AB=2\times\frac{3\sqrt{14}}{2}=3\sqrt{14}\ \text{cm}。$
答:弦$AB$的長(zhǎng)為$3\sqrt{14}\ \text{cm}。$