We start the dragon evolution with a V-Seahorse and not with a V-polygon. In this way we avoid changes of the neck vertices after several single modifications. We need a seahorse with wide head, otherwise intersections would occur in step 04.
Here we are not forced to use an angle ε = 180°/k, because the resulting V-Dragon will not admit spiral or rotational tilings of the plane. Suitable angles are from the interval [12°,18°], but in principle any angle greater than 0° can be used if it is small enough that no intersections of the circular arcs occur (this depends on the position of point A₁).

For our sample we use a V-Seahorse with wide head and internal angle ε = 15° at point Z. A good choice for the internal angle at point T is 180°-2ε, because then the angle at V is equal to ε.
Keep the point A₁ on the left ray from T variable. This allows changes the size of the head and the thickness of the neck.

Construct V-Seahorse 2 and 3 by two transformations of the original tile 1. Namely refelction of tile 1 at point S and rotation of tile 1 around Z by ε. Tile 1 is incompletely surrounded by tile 2 and 3.

Our goal is to construct a patch topologically equivant to a disc and formed by three congruent tiles, where tile 1 is surrounded by the tiles 2 and 3 Additionally 3 conditions have to be met:

Conditions

(1) Instead of only point V, tile 2 and 3 have a line segment in common near the bottom of the patch.
(2) Instead of only point Z, tile 2 and 3 have a line segment in common near the top of the patch.
(3) Couples of tile 1 and 2 admit a tiling of the plane.

First we consider point V that belongs to all three tiles. Correctly this is point V of tile 1, point Z of tile 2 and point W of tile 3. By adding the green rectangle to tile 3 we achieve that tile 2 and 3 have a line segment in common. The point V₁ does not belong to the boundary of patch any more. Let F₃ be the upper left point of the rectangle. The position of F₃ on V₃V₁ shall be variable.

As the three tiles have to be congruent, we need to add the rectangle also to the 'feet' of tile 1 and 2. As the tile 2 is upside down its foot is near the top of the patch.

The rectangle added to tile 1 cuts off a rectangle in the head of tile 2 and vice versa. This cut also has to occur in the head of tile 3. We obtain F₃' by reflecting F₃ at point R. The new rectangle with F₃' as lower left corner produces a hole in the patch. We have to fill this hole by adding a rectangle to the head of tile 1. In the following step we write H₁ instead of F₃' because this point belongs to the head of tile 1.

The rectangle added to the head of tile 1 also has to be added to the heads of tile 3 (top of the patch) and tile 2 (bottom of the patch). We obtain H₂ by reflection of H₁ at point S and H₃ by rotation of H₁ around Z by ε.
Now all tiles are congruent and the patch meets condition 1.
Normally there is a gap between the two rectangles on the bottom of the patch.

In general the gap described in step 06 prevents a tiling of the plane by the tile couple 1 and 2. To resolve this problem we can close the gap by using a length for the rectangle that is equal to the length of TZ. This may lead to overlapings of the tiles. Then we have to adjust F₃ and A₁.

After adjusting the variable points the patch looks like this:

Now also condition 3 is satisfied, because couples of tile 1 and 2 can be translated horizontally and can form a stripe of infinite length between parallel straight lines. These stripes admit a tiling of the plane in infinitely many different ways.

This is the most important step. Choose a point G₁ on the line segment H₁Z and cut off the green part of the nose of tile 1 (the angle 2ε at G₁ is suggested, but variations are possible).
Next paste the green triangle at tile 2 without changing the position of the triangle. That means the area of the patch is not changed.
Tile has now a hook at its tail (take in account that tile 2 is oriented upside down).

Of course we have to cut of the same tringle from tile 2 and 3. The triangle cutted from tile 2 is pasted to tile 1 und thus no hole occurs in the patch.

The triangle cutted from the nose of tile 3 has to be moved and pasted correctly to the tail of the same tile. Afterwards all 3 tiles are congruent again.
Now also condition 2 is satisfied and we are done.