Question

如图12,已知PA切⊙O于A,割线PBC交⊙O于B、C,PD⊥AB于D,延长PD交AO的延长线于E,连结CE并延长交⊙O于F,连结AF.

图12

(1)求证:PD·PE =PB·PC;

(2)求证:PE∥AF;

(3)连结AC,若AE∶AC=1∶,AB=2,求EF的长.

Answer 1

思路分析:(1)证明等积式往往考虑相似三角形,但△PBD与△PEC不相似,因此要用PA²=PB·PC进行等积变换.?

(2)要证明PE∥AF,只需证明同位角∠PEC和∠F相等.?

(3)首先找出EF与AB的关系,同时注意到AE∶AC=1∶,因此,先设法求出EF∶AB,这可由相似三角形得出.

(1)证明:∵PA切⊙O于A,?

∴PA²=PB·PC,PA⊥AE.?

又AD⊥PE,∴△APE∽△DPA.?

∴PA²=PD·PE.∴PD·PE =PB·PC.

(2)证明:∵PD·PE =PB·PC,∴=.?

又∠EPC =∠BPD,∴△BPD∽△EPC.?

∴∠PBD =∠PEC.又∵∠PBD =∠F,?

∴∠PEC =∠F.∴PE∥AF.

(3)解:∵PA切⊙O于A,∴∠BAP =∠ACP.?

∵∠APB =∠CPA,∴△APB∽△CPA.?

∴=.?

又∵∠ABP =∠F,∠BAP =∠AEP =∠FAE,?

∴△AEF∽△APB.∴=.?

∴=.∴= =.?

又AB =2,∴.