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先看人教版数学教材九年级上册第69页“拓广探索”第8题:如图1,过菱形对角线交点的一条直线,把菱形分成了两个梯形,这两个梯形是全等的吗?为什么?为了叙述问题的方便,不妨设菱形的四个顶点分别为A、B、C、D,菱形的两条对角线相交于点O,EF为过点O的一条直线,分别交AB、CD于点E、F于是问题转化为判断梯形AEFD与梯形CFEB是否全等.思路一利用全等图形的定义,容易证得梯形AEFD与梯形CFEB的四条边和四个内角分别对应相等.请读者朋友完成证明过程.思路二利用中心对称的定义,看看梯形AEFD与梯形CFEB是否关于点O对称.在菱形ABCD中,AO=CO,BO=DO,故点A与点
Look at the people’s education version of mathematics textbooks ninth grade on the first page 69 “to explore” the eighth question: as shown in Figure 1, the diagonal diamond crossing a straight line, the diamond is divided into two trapezoidal, the two trapezoidal Is it congruent? Why? In order to describe the convenience of the problem, may wish to set the rhombus of the four vertices were A, B, C, D, rhombus two diagonals intersect at point O, EF is o Straight line, respectively, to pay AB, CD at point E, F Then the problem is converted to judge the trapezoidal AEFD and trapezoidal CFEB is equal to the idea of using the definition of congruent graphics, trapezoidal AEFD and Trapezoidal CFEB easy to prove the four sides and four corners Respectively, corresponding to the same .Remember a friend to complete the proof process .Thinker two use the center of symmetry to see if trapezoidal AEFD and trapezoidal CFEB is symmetrical about the point O. In the diamond ABCD, AO = CO, BO = DO, the point A and point