As shown in the figure, AB is the diameter of ⊥ o, P is a point on the extension line of AB, PC is the tangent of ⊥ o, C is the tangent, BD ⊥ PC, Foot drop is D, cross o to e, connect AC.BC (1) BC ^ 2 = BD * ba. (2) if AC = 6, de = 4, find PC

As shown in the figure, AB is the diameter of ⊥ o, P is a point on the extension line of AB, PC is the tangent of ⊥ o, C is the tangent, BD ⊥ PC, Foot drop is D, cross o to e, connect AC.BC (1) BC ^ 2 = BD * ba. (2) if AC = 6, de = 4, find PC

(1) It can be proved that the triangle DBC is similar to the triangle CBA in that DB: BC = CB: Ba, = > BC ^ 2 = BD * BA (2) connects AE, then CF is perpendicular to AE and AE intersects with F through point C. Because CF is perpendicular to AE, EC = AC = 6, and because