用代数法证明帕斯卡定理请帮忙给出正确答案和分析 谢谢！

正确答案：×
适当选取射影坐标系,使简单六点形之六个顶点的坐标为A1(1,0,0),A2(0,1,0),A3(0,0,1),A4(1,1,1),A5(a1,a2,a3),A6(b1,b2,b3)．经过计算可得通过A1,A2,A3,A4,A5的二阶曲线方程为：a3(a1－a2)χ1χ2＋a1(a2－a3)χ2χ3＋a2(a3－a1)χ3χ1＝0(1)因为A1A2A3A4A5A6为二阶曲线内接简单六点形,所以有a3(a1－a2)b1b2＋a1(a2－a3)b2b3＋a2(a3－a1)b3b1＝0(2)现在计算各边方程和三对对边的交点：A1A2：χ3＝0,A2A3：χ1＝0,A3A4：χ1－χ2＝0,A4A5：(a3－a2)χ1＋(a1－a3)χ2＋(a2－a1)χ3＝0,A5A6：(a2b3－a3b2)χ1＋(a3b1－a1b3)χ2＋(a1b2－a2b1)χ＝0,A6A1：b3χ2－b2χ3＝0．设A1A2×A4A5＝L,A2A3×A5A6＝M,A3A4×A6A1＝N,于是得到：L(a1－a3,a2－a3,0),M(0,a1b2－a2b1,a1b3－a3b1,a1b3－a3b1),N(3b1),N(b2,b2,b3)．因为＝b3(a1－a3)(a1b2－a2b1)＋b2(a3－a2)(b1a3－a1b3)＋b2(a1－a3)(b1a3－a3b1)＝a3(a1－a2)b1b2＋a1(a2－a3)b2b3＋a2(a3－a1)b1b3＝0,所以L,M,N三点共线．反之,当L,M,N共线时,△＝0,即(2)成立,亦即A6在A1,A2,A3,A4,A5五点所确定的二阶曲线上．A1A2A3A4A5A6是二阶曲线的内接六点形,综上所述帕斯卡定理及其逆定理都成立．