1. 如图,在以点O为圆心的两个同心圆中,大圆的弦AB交小圆于点C,D.
(1)求证:$AC=BD$;
(2)若大圆的半径$R=10$,小圆的半径$r=8$,且圆心O到直线AB的距离为6,求AC的长.

1.(1)证明:如答图,过点O作$OE⊥AB$于点E,则$CE=DE,AE=BE$,
$\therefore AE-CE=BE-DE$,即$AC=BD$.
(2)解:由(1)可知,$OE⊥AB$且$OE⊥CD$,连接OC,OA,如答图,$\therefore OE=6$,
$\therefore CE = \sqrt{OC^2-OE^2} = \sqrt{8^2-6^2} = 2\sqrt{7}$, $AE = \sqrt{OA^2-OE^2} = \sqrt{10^2-6^2}=8$,
$\therefore AC=AE-CE=8-2\sqrt{7}$.
第1题答图

2. (2025·宜兴期末)如图,在$\odot O$中,$AB$是$\odot O$的直径,弦$CD$与$AB$交于点$E$,且$D$是$\overset{\frown}{AB}$的中点,连接$AC,BC,AD,BD$. 若$OE=2,DE=4$,求$CD$的长.

2.解:如答图,过点O作$OM⊥CD$于点M,连接OD,
$\therefore CD=2DM$.
第2题答图
$\because AB$是$\odot O$的直径,$D$是$\overset{\frown}{AB}$的中点,
$\therefore ∠ADB=∠AOD=∠BOD=90°$.
$\because OE=2,DE=4$,
$\therefore OD=\sqrt{DE^2-OE^2}=\sqrt{4^2-2^2}=2\sqrt{3}$,
由等面积法可知$S_{△ ODE}=\frac{1}{2}DE· OM=\frac{1}{2}OD· OE$,
即$4OM=2\sqrt{3}×2$,解得$OM=\sqrt{3}$,
$\therefore DM=\sqrt{OD^2-OM^2}=\sqrt{(2\sqrt{3})^2-(\sqrt{3})^2}=3$,
$\therefore CD=6$.

3. (2025·邗江区三模)如图,$\odot O$是$△ ABC$的外接圆,$∠ BAC$的平分线与$\odot O$相交于点$D$,过点$D$作直线$DE // BC$,连接$OB$,$OC$,$BD$,$CD$.
(1)判断直线$DE$与$\odot O$的位置关系,并说明理由;
(2)若$BD=2\sqrt{5}$,$BC=8$,求四边形$OBDC$的面积.

3.解:(1)直线$DE$与$\odot O$相切.
理由:如答图,连接$OD$,设$OD$与$BC$相交于点$H$,
$\because AD$平分$∠BAC$,$\therefore ∠BAD=∠CAD$,
$\therefore ∠BOD=∠COD$.
$\because OB=OC$,$\therefore OD⊥BC$,$\therefore ∠OHC=90°$.
又$\because DE// BC$,$\therefore ∠ODE=∠OHC=90°$,$\therefore OD⊥DE$.
$\because OD$是$\odot O$的半径,
$\therefore$直线$DE$与$\odot O$相切.
第3题答图
(2)由(1)知$OD⊥BC$,
$\because OB=OC$,$\therefore OD$垂直平分$BC$,$\therefore BH=\frac{1}{2}BC=4$,
$\therefore DH=\sqrt{BD^2-BH^2}=\sqrt{(2\sqrt{5})^2-4^2}=2$,
$\therefore OH=OD-DH=OB-2$.
在$Rt△ OBH$中,$OB^2=BH^2+OH^2$,
即$OB^2=4^2+(OB-2)^2$,解得$OB=5$,$\therefore OD=OB=5$,
$\therefore S_{四边形OBDC}=S_{△ OBC}+S_{△ DBC}=\frac{1}{2}BC· OH+\frac{1}{2}BC· DH=\frac{1}{2}BC(OH+DH)=\frac{1}{2}BC· OD=\frac{1}{2}×8×5=20$,
即四边形$OBDC$的面积为20.