reserve SAS for Semi_Affine_Space;
reserve a,a9,a1,a2,a3,a4,b,b9,c,c9,d,d9,d1,d2,o,p,p1,p2,q,r,r1,r2,s,x, y,t,z
  for Element of SAS;

theorem Th45:
  parallelogram a,b,c,d1 & parallelogram a,b,c,d2 implies d1=d2
proof
  assume that
A1: parallelogram a,b,c,d1 and
A2: parallelogram a,b,c,d2;
  not b,c,a are_collinear by A1,Th38;
  then
A3: not b,c // b,a;
  a,c // b,d2 by A2;
  then
A4: c,a // b,d2 by Th6;
  a,b // c,d2 by A2;
  then
A5: b,a // c,d2 by Th6;
  a,c // b,d1 by A1;
  then
A6: c,a // b,d1 by Th6;
  a,b // c,d1 by A1;
  then
A7: b,a // c,d1 by Th6;
  b,c // c,b by Def1;
  hence thesis by A3,A7,A5,A6,A4,Th19;
end;
