Question

如图2-24,△ABC内接于⊙O,AE切⊙O于点A,BD平分∠ABC,交⊙O于点D,交AF的延长线于点E,DF⊥AE于点F.

图2-24

求证:(1) =;

(2)AC =2AF.

Answer 1

思路分析:(1)运用弦切角定理证△ABE∽△DAE即可;(2)依据圆周角定理得点D为的中点.结合条件容易得到AD平分∠EAC,从而联系“角平分线上的点到角两边距离相等”,过点D作DH⊥AC,垂足为H,结合垂径定理得结论.

证明:(1)∵AE切⊙O于点A,∴∠EAD =EBA.?

又∠E =∠E,∴△DAE∽△ABE.?

∴=,即=.?

(2)过点D作DH⊥AC,垂足为H.?

∵∠EAD= ∠ABD,∠DAC= ∠DBC,BD平分∠ABC,?

∴∠EAD =∠DAC.又∵DF⊥AE,DH⊥AC,?

∴DF =DH.在Rt△DFA和Rt△DHA中,?

DF =DH,DA= DA,?

∴Rt△DFA≌Rt△DHA.∴AF =AH.?

∵ =,DH⊥AC,?

∴AH =CH,AC =2AH =2AF.