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题1已知:如图1,直线AB与⊙O相切于点C,AO交⊙O于点D,连结CD,OC.求证:∠ACD=1/2∠COD.原解如图1,作OE⊥CD于点E,则∠COE+∠OCE=90°.∵⊙O与AB相切于点C,∴OC⊥AB,即∠ACD+∠OCE=90°.∴∠ACD=∠COE.∵△ODC是等腰三角形,OE⊥CD,
Problem 1 is known: As shown in Figure 1, the line AB and ⊙O tangential to point C, AO intersect ⊙ O at point D, link CD, OC. For OE⊥CD at point E, then ∠COE + ∠OCE = 90 °. ∵O and AB tangent to point C, ∴OC⊥AB, that ∠ACD + ∠OCE = 90 ° .ACD = ∠COE.∵ △ ODC is isosceles triangle, OE⊥CD,