Question

如圖2-2-11,已知四邊形ABCD內(nèi)接于⊙O,AB是直徑,AD =DC,分別延長BA、CD交于點E,BF⊥EC,交EC的延長線于F,若EA =AO,BC =12.求CF的長.

圖2-2-11

Answer 1

思路分析:在Rt△CFB中,已知BC=12,要求CF,只有尋找與它相似的三角形,根據(jù)四邊形ABCD內(nèi)接于⊙O,則∠BCF =∠BAD,因此連結(jié)BD,構(gòu)造Rt△BAD,下面證明△BAD∽△BCF.

解:連結(jié)OD、BD,?

∵AD = DC,∴AD =DC.?

∴∠ABC  m     m      m    ∠AOD.?

∴OD∥BC.∴=.?

∵EA =AO =BO,BC=12,?

∴OD =8.∴AB =16,EB =24.?

∵四邊形ABCD內(nèi)接于⊙O,?

∴∠EDA =∠EBC.∴△EDA∽△EBC.?

∴= =.?

設(shè)AD =CD =x,ED =y,?

則= =,解得, ,?

∴.?

∵AB為⊙O的直徑,∴∠ADB =∠F =90°.?

又∠DAB =∠FCB,∴Rt△ADB∽Rt△CFB.?

∴=,即=,?

∴.