1问题的由来有这样一道题:如图,⊙O是等边△A1A2A3的外接圆,P是弧A1A3上任一点.求证:PA1+PA3=PA2.分析(1)要证明两条线段的和等于第三条线段,常用的方法有:将较长线段分成两段,然后再证明分成的两条线段与原两条线段分别对应相等;将两条较短线段合并成一条线段,再证明合并后的线段与较长线段相等.(2)由于结论中所涉及三条线段的四个顶点都在同一个圆上,且△A1A2A3是等边三角形,可以尝试用托勒密定理来证明. 1 The origin of the problem is such a question: As shown in the figure, ⊙O is the circumcircle of the equilateral △A1A2A3, P is any point on the arc A1A3. Proof: PA1+PA3=PA2. Analysis (1) To prove that the sum of the two line segments is equal to The third line segment, commonly used methods are: the long line segment is divided into two sections, and then proved to be divided into two line segments and the original two line segments corresponding to equal; the two short line segments merged into a line segment, and then prove the merger The line segment is equal to the long line segment. (2) Since the four vertices of the three line segments involved in the conclusion are all on the same circle, and ΔA1A2A3 is an equilateral triangle, it can be proved by Ptolemy’s theorem.