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近年来,在各地中考试卷中,经常可见比较三条线段之间的大小关系或平方关系的一类试题,而三条线段往往较为分散.应对这类试题的一个有效方法是将涉及这三条线段之一的某个图形旋转一个角度,使分散的线段集中到一个三角形里,然后利用三角形边之间的不等关系或勾股定理进行解答.例1(2008年天津)已知Rt△ABC中,∠ACB=90°,CA=CB,有一个圆心角为45°,半径的长等于CA的扇形CEF绕点C旋转,且直线CE,CF分别与直线AB交于点M,N.(1)当扇形CEF绕点C在∠ACB的内部旋转时,如图1,求证:MN2=AM2+BN2;(2)当扇形CEF绕点C旋转至图2的位置时,关系式MN2=AM2+BN2是否仍然成立?若成立,请证明;若不成立,请
In recent years, exam papers around the world often can be seen comparing the size of the relationship between the three lines or the relationship between the square of a class of questions, and the three lines are often more dispersed.To deal with such questions is an effective way to involve one of these three lines Of a graph is rotated by an angle, the scattered segments concentrated in a triangle, and then use the unequal relationship between the triangle sides or the Pythagorean theorem to answer. Example 1 (Tianjin 2008) Rt △ ABC, ∠ ACB = 90 °, CA = CB, with a central angle of 45 ° and a fan-shaped CEF with a radius equal to CA rotates around point C, and the lines CE and CF intersect the straight line AB at points M and N. (1) When the fan-shaped CEF is rotated around the inside of ∠ACB, as shown in Figure 1, verify that: MN2 = AM2 + BN2; (2) When the fan-shaped CEF rotates about point C to the position of Figure 2, Still valid? If yes, please prove; if not, please