【教材变式】(教材$P_{102}T_{12}$变式)已知AB是⊙O的直径,$\overset{\frown}{AD}= \overset{\frown}{DE}$,BE⊥CD于点C.
(1)如图1,若CE= 1,CD= 2,求AB的长；

(2)如图2,若CE= 4,BE= 6,求OC的长；
(3)如图3,若CD= 4,BE= 6,求BD的长；

(4)如图4,若AB= 10,CD= 4,求AC的长.

解:(1)连接$OD$,$AE$交于点$F$.
$\because AB$是$\odot O$的直径,
$\therefore \angle AEB=\angle AEC=90^{\circ}$.
$\because \overset{\frown}{AD}=\overset{\frown}{DE}$,
由圆的对称性得$OD\perp AE$,
$AF=EF$.
又$\because BE\perp CD$,
$\therefore$四边形$CDFE$为矩形,
$\therefore DF=CE=1$,$CD=EF=2$.
设$OF=x$,则$OA=OD=x+1$.
在$\text{Rt}\triangle AOF$中,
$(x+1)^{2}=2^{2}+x^{2}$,解得$x=\dfrac{3}{2}$,
$\therefore OA=\dfrac{5}{2}$,$AB=2OA=5$;
(2)连接$OD$,过点$O$作$OH\perp BC$于点$H$.
$\because BE=6$,$\therefore EH=BH=3$,
由(1)知$OD\perp CD$.又$\because BE\perp CD$,
$\therefore$四边形$OHCD$为矩形,
$\therefore OD=CH=7$,在$\text{Rt}\triangle OBH$中,
$OH^{2}=OB^{2}-BH^{2}=7^{2}-3^{2}=40$.
在$\text{Rt}\triangle OCH$中,$OC^{2}=OH^{2}+CH^{2}=40+49=89$,$\therefore OC=\sqrt{89}$;
(3)连接$OD$,过点$O$作$OH\perp BE$于点$H$.
$\because BE=6$,$\therefore EH=BH=3$,
由(1)知$OD\perp CD$.又$\because BE\perp CD$,
$\therefore$四边形$OHCD$为矩形,
$\therefore OH=CD=4$.在$\text{Rt}\triangle BOH$中,
$OB=\sqrt{OH^{2}+BH^{2}}=\sqrt{4^{2}+3^{2}}=5$,
$\therefore CH=OD=OB=5$,$\therefore BC=8$.
在$\text{Rt}\triangle BCD$中,
$BD=\sqrt{CD^{2}+BC^{2}}=\sqrt{4^{2}+8^{2}}=4\sqrt{5}$;
(4)连接$OD$,$AE$交于点$F$,由(1)知四边形$CDFE$为矩形,$AF=EF$,
$\therefore AE=2EF=2CD=8$.
$\because AB=10$,
$\therefore BE=\sqrt{AB^{2}-AE^{2}}=6$.
$\because AF=EF$,$AO=BO$,
$\therefore OF=\dfrac{1}{2}BE=3$.
$\because OD=\dfrac{1}{2}AB=5$,
$\therefore DF=CE=5-3=2$.
在$\text{Rt}\triangle AEC$中,
$AC=\sqrt{AE^{2}+CE^{2}}=\sqrt{64+4}=2\sqrt{17}$.