解：連接?$OB.$?
∵?$AC$?是?$\odot O$?的直徑，弦?$BD\perp AO$?于點(diǎn)?$E，$??$BD = 8\ \mathrm {cm}，$?
∴?$BE=\frac {1}{2}BD = 4\ \mathrm {cm}.$?
設(shè)?$\odot O$?的半徑為?$x\mathrm {cm}，$?
則?$OB = OA = x\mathrm {cm}，$??$OE=(x - 2)\mathrm {cm}.$?
在?$Rt\triangle OEB$?中，由勾股定理，得?$OE^2+BE^2=OB^2，$?
即?$(x - 2)^2+4^2=x^2，$?
?$ $?展開式子得?$x^2-4x + 4+16=x^2，$?
?$ $?移項(xiàng)可得?$-4x=-20，$?
?$ $?解得?$x = 5.$?
∴?$\odot O$?的半徑為?$5\ \mathrm {cm}，$?
∴?$EC=2×5 - 2 = 8(\mathrm {cm})，$?
∴在?$Rt\triangle BEC$?中，?$BC=\sqrt {BE^2+EC^2}=\sqrt {4^2+8^2} = 4\sqrt {5}(\mathrm {cm}).$?
∵?$OF\perp BC，$??$OF{過圓心}，$?
∴?$CF=\frac {1}{2}BC = 2\sqrt {5}\mathrm {cm}，$?
∴在?$Rt\triangle OFC$?中，?$OF=\sqrt {OC^2-CF^2}=\sqrt {5^2-(2\sqrt {5})^2}=\sqrt {25 - 20}=\sqrt {5}(\mathrm {cm})$?
