So EF//AB, EF=AB/2.

In the same way; In a similar way

HG//AB，GH=AB/2

therefore

HG//EF, and HG=EF.

So the quadrilateral EFGH is a parallelogram.

Must be on the same plane.

It can also be obtained from the neutral line.

Er //CD

therefore

CD// airplane EFGH

In the space quadrilateral ABCD, two opposite sides AB=CD=3, E and F are points on AD and BC respectively, AE∶ED=BF∶FC= 1∶2, EF= the angle formed by straight line AB and CD.

Solution: As shown in Figure 2, connect BD into EG‖AB,

Cross BD at G point and connect GF.

∴AE∶ED=BG∶GD。

Figure 2

AE∶ED=BF∶FC，

Therefore, BG∶GD=BF∶FC.

∴ GF‖CD .

∴∠EGF (or complementary angle) is the angle formed by non-planar straight line AB and CD.

In δδEGF, EG= AB=2, GF= CD= 1, EF=.

According to the cosine theorem, there is cos∠EGF= =-

∴∠EGF= 120

Therefore, the angle formed by the non-planar straight line AB and CD is 60.