1986年数学高考第三题即课本145页第二题为: 如图1,AB是圆O的直径,PA垂直于圆O所在的平面,C是圆周上的任意点.求证:△PAC所在的平面垂直于△PBC所在的平面. 该题粗看似乎没有几种证法;但实际上证明的渠道较多.尽管有的证法并不简便,但作为思路,自有其参考价值. The third question of the 1986 mathematics college entrance examination is the textbook 145 pages of the second question is: Figure 1, AB is the diameter of the circle O, PA is perpendicular to the plane where the circle O, C is an arbitrary point on the circumference. Proof: PAC where The plane is perpendicular to the plane in which △PBC is located. There does not seem to be any kind of proof in the rough view of this question; but there are actually more channels to prove. Although some proofs are not simple, they have their own reference value as a train of thought.