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利用旋转解题,往往是在有相等边的特殊图形中,比如正方形、菱形、等边三角形、等腰直角三角形等.下面举例子说明.例1如图1,△ABC中,∠ACB=90°,AC=BC,P是△ABC内一点,且PA=3,PC=2,PB=1.试求∠CPB的度数.分析:在这个问题中,已知线段PA、PB、PC没有明显的特殊位置关系,不论把PA截成两段还是把PC(PB)延长使其与PA相等,对解题均无明显帮助.因此,考虑到△ABC的特殊性,可以把∠CPB所在的△PBC绕C点逆时针旋转90°,则点B与点A重合,如图2,点P旋转到点P’的位置,已知的三条线段PA、
The use of rotation problem solving, often in the special graphics with equal sides, such as square, rhombus, equilateral triangle, isosceles right triangle, etc. Example 1 as shown in Figure 1, △ ABC, ∠ ACB = 90 °, AC = BC, P is a point within △ ABC, and PA = 3, PC = 2, PB = 1. The degree of CPB is sought. Analysis: In this problem, Of the special location relationship, regardless of the PA cut into two sections or extend the PC (PB) to equal PA, no significant help to solve the problem.Therefore, taking into account the special nature of △ ABC, you can put ∠ CPB where △ PBC around the C point counterclockwise rotation of 90 °, the point B and point A coincide, as shown in Figure 2, point P to the point P ’position, the known three lines PA,