解：过点

D

作

D M ⊥ B C

于

M

，连接

O B

．

∵ $A H$ 为 $B C$ 的垂直平分线，

∴ $B H = H C = \frac{1}{2} B C = 3$ ．

∴ $O H = \sqrt{O B^{2} - B H^{2}} = \sqrt{5^{2} - 3^{2}} = 4$ ．∴ $A H = O A + O H = 5 + 4 = 9$ ．

∴ $A B = A C = \sqrt{A H^{2} + C H^{2}} = \sqrt{9^{2} + 3^{2}} = 3 \sqrt{10}$ ．

∴ $C D = \frac{1}{2} A C = \frac{3}{2} \sqrt{10}$ ．

∵ $A H ⊥ B C ， D M ⊥ B C$ ，

∴ $D M ∥ A H$

∴ $△ C M D ∽ △ C H A$ ，

又 $A D = C D$ ，

∴ $\frac{D M}{A H} = \frac{C M}{C H} = \frac{C D}{C A} = \frac{1}{2}$ ．

∴ $M H = \frac{1}{2} H C = \frac{3}{2}$ ， $D M = \frac{1}{2} A H = \frac{9}{2}$ ．

∴ $B M = B H + M H = 3 + \frac{3}{2} = \frac{9}{2}$ ．

∴ $B D = \sqrt{B M^{2} + D M^{2}} = \sqrt{{(\frac{9}{2})}^{2} + {(\frac{9}{2})}^{2}} = \frac{9}{2} \sqrt{2}$ ．

∵ $∠ C F D = ∠ B A D ， ∠ F D C = ∠ A D B$ ，

∴ $△ F C D ∽ △ A B D$ ．

∴ $\frac{F C}{A B} = \frac{C D}{B D}$ ．

∴ $\frac{F C}{3 \sqrt{10}} = \frac{\frac{3}{2} \sqrt{10}}{\frac{9}{2} \sqrt{2}}$ ．

∴ $F C = 5 \sqrt{2}$ ．

湖北省潜江、天门、仙桃、江汉油田2023年中考数学试卷