24. (13分)如图,$\triangle ABC$内接于$\odot O,AB$是$\odot O$的直径$,CD$平分$\angle ACB$交$\odot O$于点$D$,连接$BD$,过点$C$作$\odot O$的切线,交$AB$的延长线于点$E$.
(1) 求证:$\angle ECB=\angle EAC$;
(2) 若$BD=4\sqrt{2},BE=3$,求$CE$的长.
(1) 求证:$\angle ECB=\angle EAC$;
(2) 若$BD=4\sqrt{2},BE=3$,求$CE$的长.
答案：
24.(1)如图，连接OC.
∵CE是⊙O的切线，
∴OC⊥CE.
∴∠ECB + ∠OCB = 90°.
∵AB是⊙O的直径，
∴∠ACB = 90°.
∴∠EAC + ∠OBC = 90°.
∵OB = OC,
∴∠OBC = ∠OCB.
∴∠ECB = ∠EAC (2)如图,连接AD.
∵CD平分∠ACB,
∴∠ACD = ∠BCD.
∴$\overset{\frown}{AD} = \overset{\frown}{BD}.$
∴AD = BD = 4√{2}.
∵AB是⊙O的直径,
∴∠ADB = 90°.
∴AB = √{BD² + AD²} = 8.
∴EA = AB + BE = 11.
∵∠ECB = ∠EAC,∠E = ∠E,
∴△ECB∽△EAC.
∴$\frac{EB}{EC} = \frac{EC}{EA},$即$\frac{3}{EC} = \frac{EC}{11}.$
∴EC = √{33}
24.(1)如图，连接OC.
∵CE是⊙O的切线，
∴OC⊥CE.
∴∠ECB + ∠OCB = 90°.
∵AB是⊙O的直径，
∴∠ACB = 90°.
∴∠EAC + ∠OBC = 90°.
∵OB = OC,
∴∠OBC = ∠OCB.
∴∠ECB = ∠EAC (2)如图,连接AD.
∵CD平分∠ACB,
∴∠ACD = ∠BCD.
∴$\overset{\frown}{AD} = \overset{\frown}{BD}.$
∴AD = BD = 4√{2}.
∵AB是⊙O的直径,
∴∠ADB = 90°.
∴AB = √{BD² + AD²} = 8.
∴EA = AB + BE = 11.
∵∠ECB = ∠EAC,∠E = ∠E,
∴△ECB∽△EAC.
∴$\frac{EB}{EC} = \frac{EC}{EA},$即$\frac{3}{EC} = \frac{EC}{11}.$
∴EC = √{33}