Question

如图2-5-12,已知AB是⊙O的直径,CA交弦BF延长线于E,DE⊥AC于E,CB交⊙O于D且AB=AC,求证:AE·EC=BE·EF.

图2-5-12

Answer 1

证明:连结OD、AD,∵AB是⊙O的直径,

∴∠ADB=90°.∴AD=BC.

∵AB=AC,∴BD=DC.

∵BO=OA,∴OD∥AC.

∵DE⊥AC,∴DE⊥OD.

∴DE是⊙O切线,∴DE²=EF·EB.①

在Rt△ACD中,DE⊥AC,

∴DE²=AE·EC.②

∴由①②得AE·EC=BE·EF.