解：如圖?$.$?設(shè)?$BE = t.$?
∵?$EF = 10，$?
∴?$OE = OG = OH = 5.$?
∵?$∠GOH = 90°，$?
∴?$∠AOG+∠BOH = 90°.$?
∵在矩形?$ABCD$?中，?$∠DAB=∠ABC = 90°，$?
∴?$∠AGO+∠AOG = 90°，$?
∴?$∠AGO=∠BOH.$?
在?$\triangle GAO$?和?$\triangle OBH$?中，
?$\begin {cases}∠GAO=∠OBH = 90°，\\∠AGO=∠BOH，\\OG = HO，\end {cases} $?
∴?$\triangle GAO\cong \triangle OBH，$?
∴?$GA = OB = BE - OE=t - 5.$?
∵?$AB = 7，$?
∴?$AE = BE - AB=t - 7，$?
∴?$AO = OE - AE = 5-(t - 7)=12 - t.$?
在?$\text{Rt}\triangle GAO$?中，由勾股定理，得?$AG^2+AO^2=OG^2，$?
∴?$(t - 5)^2+(12 - t)^2=5^2，$?
即?$t^2-17t + 72 = 0，$?
解得?$t_{1}=8，$??$t_{2}=9，$?
∴?$BE$?的長(zhǎng)為?$8$?或?$9$?
