Proving a quadrilateral may have at most four axes of symmetry

Question

I've attempted to construct a proof that a non-self-intersecting quadrilateral $\square ABCD$ can have at most four axes of symmetry:

There must be the same number of vertices on either side of the axis, it is impossible to have vertices on either side since then the axis would necessarily pass through all four vertices which is impossible since they aren't collinear.

Case 1 - Two vertices on either side of the axis:

If the axis intersects a side not at one of its vertices, it must intersect the midpoint of that side for the two segements to be symmetric about the axis. I'm really stuck here as to a simple argument, though I can show that a line can only pass through two midpoints, I'm not sure if it helps

Case 2 - One vertex on either side of the axis:

The axis must pass through two of the vertices so that only two vertices remain. The axis must pass through a pair of non-adjacent vertices because if it passed through two adjacent vertices, it would be one of the sides of the quadrilateral. This would leave three sides of the quadrilateral on one side of the axis and none on the other which is not symmetrical. Hence, since there are two pairs of non-adjacent vertices, there are two axes of symmetry for this case.

As you can see the proof is incomplete and I need guidance on how to make it simpler and how to finish it. I expect there is a very simple argument for this.

Answer 1

A square has 4 axes of symmetry (one horizontal, one vertical, and two diagonal). A rectangle or a rhombus that is not a square has 2 axes of symmetry (one horizontal and one vertical, or two diagonal, respectively). An isosceles trapezoid that is not a rectangle, or a kite that is not a rhombus, has 1 axis of symmetry. Finally, all other simple quadrilaterals have no axes of symmetry. In all cases, there are at most 4 axes of symmetry.

Answer 2

Propostion: An axe of symmetry divide a polygon in two same parts.

Corolario: An axe of symmetry divide the plane in two parts. Each part has the same amount of vertex from the polygon.

If you have this, then you have the two options that you were talking about.

Case 1: If you want two vertices in each side, the axe has to pass throught two opposite sides of the quadrilateral and divide each side in two same parts beacause the distance between two corresponding vertices is the same. Then you have two axes at most, cause you have two pair of opposite sides.

Case 2: One vertex in each side, then the axe is the diagonal. (two axes at most).

My english is not the best, but I hope I helped.