Jacobs makes a delightful use of this illusion to discuss a property of rotations. A rotation f can be determined by two arbitrary points P and Q and by their images f(P) and f(Q), provided neither is fixed. In the applet we think of E as being the image of A and H the image of D. The center of rotation O is bound to lie on the perpendicular bisectors of AE and DH and hence at their intersection. (In principle, the two two lines may coincide. Then the lines AE and DH intersect at the center of rotation.)
So assume the parallelograms ABCD and EFGH are equal and that O has been found as just described. In triangles ADO and EHO, AO = EO and DO = HO by construction, and AD = EH by our assumption. The triangles are equal by SSS. In particular, their angles at O are equal. If we add to each the angle DOE, we arrive at
AOE = DOH,
So that indeed when A is rotated around O to coincide with E, D rotates onto H. The rest of the parallelogram ABCD comes along as a rigid body.