As shown in the figure, BC is the diameter of circle O, a is a point on circle O, cross point C to make the tangent line of circle O, intersect the extension line of BA at point D, and take the midpoint e of CD, the extension line of AE and the tangent line of BC (1) prove that AP is the tangent of circle O. (2) if OC = CP, ab = 3 √ 3, find the length of CD

As shown in the figure, BC is the diameter of circle O, a is a point on circle O, cross point C to make the tangent line of circle O, intersect the extension line of BA at point D, and take the midpoint e of CD, the extension line of AE and the tangent line of BC (1) prove that AP is the tangent of circle O. (2) if OC = CP, ab = 3 √ 3, find the length of CD

(1) It is proved that if OA and OE are connected, because CD is the tangent line of circle O, then ∠ BCD = 90 ° because ed = EC, OB = OC  OE ‖ BD ∈ COE = ∠ oba ∠ Bao = ∠ AOE, because ob = OA ≠ oba = ∠ OAB ∈ COE = ∠ aoeoa = OC, OE = OE ≠ △ OCE & # 8773; △ OAE (SAS) ∠ OAE = ∠ OCE = 90 ∈ AP ⊥ OA ⊥ ap