证明正方体对角线交于一点

证明：可用解析几何方法；设正方体ABCD-A'B'C'D' 边长为a；以A为原点,AB为X轴正向；AD为Y轴正向,AA'为Z轴正向建立空间坐标系；则各点坐标为：A(0,0,0),B(a,0,0),C(a,a,0),D(0,a,0)；A'(0,0,a),B'(a,0,a),C'(a,a,a),D...