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1问题引入题1:如图1,已知AD平分∠BAC,BD⊥AD,E是BC的中点,求证:ED=1/2(AB-AC)。分析:AB-AC是用差的形式表示的一条线段,而题设中线段AD一方面平分∠BAC,另一方面又垂直于BD,由此就可以推想到AD是某一个三角形∠A的平分线,又是∠A对边上的高,这个三角形一定是以∠A为顶角的等腰三角形。于是作出如图2所示辅助线,补全图形使△ABG成为等腰三角形,从而构建解决问题的“系统”。在此系统中,以三角形的
1 Problem Introduction Question 1: As shown in Fig. 1, it is known that AD bisects ∠BAC, BD⊥AD, and E are the midpoints of BC. Verify that: ED = ½ (AB-AC). Analysis: AB-AC is the poor form of a line segment, and set the line segment AD on the one hand bisects BAC, on the other hand perpendicular to the BD, it can be assumed that AD is a triangle bisect ∠ A Line, which is 高A on the edge of the high, this triangle must be 等 A isosceles triangle vertex angle. Then make the auxiliary line as shown in Figure 2, make up the graphics so △ ABG becomes isosceles triangle, in order to build a solution to the problem “system.” In this system, it is triangular