## Symmetries on vertices of quadrilaterals

Draw an arbitrary quadrilateral ABCD. Then an arbitrary point E. Then find the points F, G, H, I as follows:
F is the symmetric of E w.r. to A, G is the symmetric of F w.r. to B, etc.
The final point I together with E and G build a triangle EGI, whose shape does not dependent on the position of E. In fact, triangle (EGI) is homothetic to (JLK), where L is the middle of BC and J, K are the middles of the diagonals. The homothety ratio is 4.
This implies that I coincides with E, exactly when J and K coincide i.e. the original quadrilateral is a parallelogram.

Free movable are :
Point E and the quadrilateral ABCD.
Switch to the selection-tool (Ctrl+1), catch and modify these shapes.

The key-fact is that an even number of point-symmetries gives a composition that is a translation (in general) or the identity.
Prove that symmetric polygons (about a center of symmetry X) give compositions which coincide with the identical transformation (look at the file : SymmetriesOnVerticesEven2.html ).

The corresponding problem for pentagons is handled in the file: SymmetriesOnVerticesOdd.html .

Are the symmetric polygons the only ones with this property? The answer is no and is handled in the file: NonSymmetricClosed.html .