Question

如圖2-4-18(1),四邊形ABCD是⊙O的內接四邊形,A是的中點,過A點的切線與CB的延長線交于點E.

           

  (1)                               (2)

圖2-4-18

(1)求證:AB·DA=CD·BE;

(2)如圖2-4-18(2),若點E在CB延長線上運動,使切線EA變為割線EFA,其他條件不變,問具備什么條件使原結論成立?

Answer 1

思路分析:(1)只需證△ABE∽△CDA.?

(2)如題圖(2),要使結論仍然成立,注意到∠ABE =∠ADC始終成立,因此仍然只需使△ABE∽△CDA即可,這樣只要另一組對應角相等即可,即只需∠BAE =∠ACD或∠E =∠CAD.

(1)證明:連結AC,∵AE切⊙O于A,?

∴∠EAB =∠ACB.?

∵ =,?

∴∠ACD =∠ACB.?

∴∠EAB =∠ACD.?

又∵四邊形ABCD內接于⊙O,?

∴∠ABE =∠CDA.∴△ABE∽△CDA.?

∴=.∴AB·DA =CD·BE.

(2)解:當 =時,∠EAB =∠ACD,

又∠ABE =∠ADC,?

∴△ABE∽△ACD,?

∴AB·DA =CD·BE,此時仍然成立.