由a1∥a2,a2⊥a3,a3∥a4,a4⊥a5,得
a1∥a5,a1∥a,6,a,1⊥a7,a1⊥a8,……
所以当下标被4除余1或2时,直线与a1平行,当下标被4整除或余3时,直线与a1垂直,
所以a1∥a2001