How to cut that shape into two congruent shapes? The shape(Roughly to say, it a 4*5 rectangle - 2*2 rectangle):

You can draw a line(polygonal chain or a curve, and may violate the grid) to cut it into two shape, for example:

But that doesn't cut it into two congruent shapes.
So my question:
Is there any way to cut that shape into two congruent shapes?
It may be trivial, but I had try to solve it for hours...
 A: Assuming the pieces do not have to be connected, we have the following solution: 
A: I think I proved that it is impossible to cut it into two congruent shapes (please correct me if there are any mistakes), under the following assumptions:

(a) The line you draw is polygonal (i.e. it consists of finitely many line segments, not necessarily horizontal/vertical, and the points don't need to have integral coordinates);
(b) The two shapes are connected;
(c) The rotation angle is a multiple of 90 degree (horizontal and vertical flip is allowed).

If these two shapes are congruent, then

(1) They have the same  circumference;
(2) They have the same number of vertices;
(3) They have the same axis-aligned bounding box, which means a minimal rectangle containing that shape, with four edges parallel to x/y axis (this holds by assumption (c)).

By (1), when the start point of the line you draw is determined, the end point is also determined. (The circumference of the whole shape is $18$, so these two points must cut it into two lines of length $9$.)
Assume the bottom-left corner has coordinate $(0,0)$, then by (2), the start points $(0,y)$, $2\leqslant y\leqslant 4$, and $(x,0)$, $2\leqslant x\leqslant 5$ are excluded.
By (3), points $(0,y)$, $0\leqslant y\leqslant 2$ are also excluded (the bounding box of the lower-right piece has a side exactly $5$, but another piece couldn't have this).
Points $(x,0)$, $0<x<1$ are also excluded (the bounding box of the lower-right piece has a side $>4$ and an edge of this piece with length $>4$ is on the boundary of the bounding box, but another piece couldn't have this).
Points $(x,0)$, $1<x<2$ are also excluded (the bounding box of the upper-left piece has a exactly $4$ and an edge of this piece with length exactly $4$ on the boundary of the bounding box, but another piece couldn't have this).
Finally we conclude that the start point must be $(1,0)$, so the end point must be $(3,3)$. But now the upper-left piece has edges $(1,0)$--$(0,0)$--$(0,4)$--$(3,4)$--$(3,3)$; you can't find the corresponding edges of the lower-right piece.
