如圖，⊙O的半徑OD，OE分別垂直于弦AB和AC，連結DE交AB，AC于F，G。求證：⑴AF^2=AB^2（已證出）⑵AG^2=DF·GE

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解: ∵OD⊥AB OE⊥AC ∴AD⌒=DB⌒ AE⌒=EC⌒ (AB⌒表示AB劣弧) ∴∠AFG=1/2×(DB⌒+AE⌒) ∠AGF=1/2×(AD⌒+EC⌒) ∴∠AFG∠AGF ∴AG=AF連AD,AE ∠DAF= 1/2×(DB⌒) ∠AEG= 1/2×(AD⌒) ∴∠DAF＝∠AEG ∠ADF= 1/2×(AE⌒) ∠EAG= 1/2×(EC⌒) ∴∠ADF＝∠EAC ∴△ADF∽△AEG ∴AF/GE=DF/AG AG×AF=DF×GE既AG^2=DF·GE