题目

如图17,已知△ABC是⊙O的内接三角形,PA是切线,PB交AC于E点,交⊙O于D点,且PE =PA,∠ABC=60°,PD=1,BD =8,则CE的长为(　　)图17A.                     B.9                    C.                    D.4 答案：思路解析:由弦切角定理得∠PAE =∠ABC =60°,又∵PE =PA,∴△PAE为等边三角形.由切割线定理得PA2=PD·PB,求得PA =3=AE =PE,∴DE =PE-PD=3-1=2;BE =BD -DE =8-2 =6.由相交弦定理得BE·ED =AE·EC.∴ = =4.答案:D

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