2013年第12期问题解答901.如图1,MN是⊙O的直径,以M为圆心的圆交⊙O于点P、Q,A是⊙M上一点(异于点P、Q),直线AP、AN、AQ分别交⊙O于点B、C、D,求证:(AB)/(AP)+(AD)/(AQ)=(NC)/(AC).(401555重庆市合川太和中学袁安全供题)证:连MQ、NQ、BC、BD、BN、PQ,则PQ⊥MN,MQ⊥GN,即QN是⊙M的切线.于是∠QAB=∠1=∠QMN=∠QDN,所以AB∥DN.又∠1+∠ABN=180°,故∠QAB+∠ABN=180°,所以AD∥BN.故四边形ABND是平行四边形,AD=BN,AB=DN. 2013 Issue 12 Q & A 901. As shown in Figure 1, MN is the diameter of ⊙O. The circle with M as its center intersects ⊙O at point P, Q, and A is the previous point (different from point P, Q) Straight line AP, AN, AQ respectively pay ⊙ O at points B, C, D, verify: AB / AP / AD / AQ = NC / AC. (401555 Chongqing Aikawa too And secondary school Yuan security question) card: even MQ, NQ, BC, BD, BN, PQ, then PQ ⊥ MN, MQ ⊥ GN, that is QN is the tangent of ⊙ M. Then ∠ QAB = ∠1 = ∠ QMN = ∠ Therefore, AB ∥ DN. And ∠ 1 + ∠ ABN = 180 °, so ∠ QAB + ∠ ABN = 180 °, so AD∥BN. Therefore, the quadrilateral ABND is a parallelogram, AD = BN, AB = DN.