题目如图1,PA为⊙O的切线,PBC为⊙O的割线,AD⊥OP于点D,△ADC的外接圆与BC的另一个交点为E.证明:∠BAE=∠ACB.这是2012年全国初中数学联合竞赛第二试(B)第二题,是一道内涵丰富、不落俗套、颇具启发性的好题,本文就此题的解法作以下探讨.1.试题原证回放如图2,连接OA、OB、OC.因为AD⊥OP,OA⊥AP, The subject is shown in Figure 1, PA is the tangent of ⊙O, PBC is the secant of ⊙O, AD⊥OP is at point D, and the intersection of the circumcircle of △ ADC and BC is E. Proof: ∠BAE = ∠ACB. Is the 2012 second national junior high school mathematics competition second test (B) second question, is a rich connotation, unconventional, quite enlightening good question, this article on the solution to the following discussion .1. Figure 2, connecting OA, OB, OC. Because AD ⊥ OP, OA ⊥ AP,