23.「2026江苏宿迁宿豫期中，★☆」(8分)如图，AB是$\odot O$的直径，CD是$\odot O$的弦，$CD ⊥ AB$于点E，点F在$\odot O$上，且$\overset{\frown}{CF}=\overset{\frown}{CA}$，连接AF.
(1)求证:$AF=CD$.
(2)连接BD，若$AE=2$，$AF=8$，求BD的长.

23.解析 (1)证明:$\because AB$是$\odot O$的直径,$CD$是$\odot O$的弦,
$CD ⊥ AB$,$\therefore \overset{\frown}{DA}=\overset{\frown}{CA}$. ……………… (2分)
$\because \overset{\frown}{CF}=\overset{\frown}{CA}$,$\therefore \overset{\frown}{CF}=\overset{\frown}{DA}$,
$\therefore \overset{\frown}{AF}=\overset{\frown}{CD}$,
$\therefore AF=CD$. ……………… (4分)
(2)如图,连接$OD$,则$OD=OA$,
$\because AF=CD$,$AF=8$,$\therefore CD=8$.
$\because AB$是$\odot O$的直径,$CD$是$\odot O$的弦,$CD ⊥ AB$于点$E$,
$\therefore CE=DE=\dfrac{1}{2}CD=4$,$∠ BED=90°$. ……………… (5分)
$\because AE=2$,$\therefore OE=OA-2$.
$\because OE^2+DE^2=OD^2$,$\therefore (OA-2)^2+4^2=OA^2$,解得$OA=5$,
$\therefore AB=2OA=10$,$\therefore BE=AB-AE=8$, ……………… (6分)
$\therefore BD=\sqrt{DE^2+BE^2}=\sqrt{4^2+8^2}=4\sqrt{5}$,
$\therefore BD$的长是$4\sqrt{5}$. ……………… (8分)