Question

如圖12,已知PA切⊙O于A,割線PBC交⊙O于B、C,PD⊥AB于D,延長PD交AO的延長線于E,連結(jié)CE并延長交⊙O于F,連結(jié)AF.

圖12

(1)求證:PD·PE =PB·PC;

(2)求證:PE∥AF;

(3)連結(jié)AC,若AE∶AC=1∶,AB=2,求EF的長.

Answer 1

思路分析:(1)證明等積式往往考慮相似三角形,但△PBD與△PEC不相似,因此要用PA²=PB·PC進(jìn)行等積變換.?

(2)要證明PE∥AF,只需證明同位角∠PEC和∠F相等.?

(3)首先找出EF與AB的關(guān)系,同時(shí)注意到AE∶AC=1∶,因此,先設(shè)法求出EF∶AB,這可由相似三角形得出.

(1)證明:∵PA切⊙O于A,?

∴PA²=PB·PC,PA⊥AE.?

又AD⊥PE,∴△APE∽△DPA.?

∴PA²=PD·PE.∴PD·PE =PB·PC.

(2)證明:∵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,∴.