Question

如图2-4-18(1),四边形ABCD是⊙O的内接四边形,A是的中点,过A点的切线与CB的延长线交于点E.

           

  (1)                               (2)

图2-4-18

(1)求证:AB·DA=CD·BE;

(2)如图2-4-18(2),若点E在CB延长线上运动,使切线EA变为割线EFA,其他条件不变,问具备什么条件使原结论成立?

Answer 1

思路分析:(1)只需证△ABE∽△CDA.?

(2)如题图(2),要使结论仍然成立,注意到∠ABE =∠ADC始终成立,因此仍然只需使△ABE∽△CDA即可,这样只要另一组对应角相等即可,即只需∠BAE =∠ACD或∠E =∠CAD.

(1)证明:连结AC,∵AE切⊙O于A,?

∴∠EAB =∠ACB.?

∵ =,?

∴∠ACD =∠ACB.?

∴∠EAB =∠ACD.?

又∵四边形ABCD内接于⊙O,?

∴∠ABE =∠CDA.∴△ABE∽△CDA.?

∴=.∴AB·DA =CD·BE.

(2)解:当 =时,∠EAB =∠ACD,

又∠ABE =∠ADC,?

∴△ABE∽△ACD,?

∴AB·DA =CD·BE,此时仍然成立.