The shaded triangle is a $(1, 2, \sqrt{5})$ right triangle. Such a triangle will have area equal to $1/5$ of the hypotenuse squared.
The square with area $S$ has side length $\sqrt{S}$, and the right triangle has a hypotenuse that is half that side length: $\frac{\sqrt{S}}{2}$. So the triangle's area is $\frac{1}{5}\left(\frac{\sqrt{S}}{2}\right)^2 = \frac{S}{20}$.
To see why a $(1, 2, \sqrt{5})$ right triangle must have an area that's 1/5 of its hypotenuse:
Suppose its legs have length $L$ and $2L$. Then the area is $A = \frac{1}{2}L \cdot 2L = L^2$, and by the pythagorean theorem its hypotenuse is $h = \sqrt{L^2 + (2L)^2} = \sqrt{5} L$. Thus, $h^2 = 5 L^2 = 5 A \implies A = \frac{1}{5}h^2$
To see that the shaded triangle is a $(1, 2, \sqrt{5})$ right triangle:
Note that it's similar to the triangle with vertices $\{A, D,$ midpoint-of-$AB\}$ or the one with vertices $\{C, B,$ midpoint-of-$CD\}$, which are each clearly $(1, 2, \sqrt{5})$ right triangles since they have one leg that's twice the length of the other.
If you can convince yourself that the shaded triangle must also be a right triangle, then the fact that one of its acute angles matches those triangles is enough to show that it's similar.
To see that shaded triangle is a right triangle, note that one of its angles is congruent with an angle of the small square in the middle (since these are vertical angles).
One way to convince yourself that the small quadrilateral in the middle really is a square (and thus all its angles are right angles) is by noting that the entire figure is unchanged by a 90 degree rotation. Thus, all four sides of the central quadrilateral must be congruent, and likewise for all four of its angles.
If you also want a way to formally show that the figure is unchanged under 90 degree rotation, consider this description which fully defines the figure:
"It is a square with vertices {A, B, C, D} and with additional line segments {A to midpoint(B,C), B to midpoint(C,D), C to midpoint(D,A), D to midpoint(A,B)}."
Now consider a rotation that takes A to A', B to B', C to C' and D to D'. After this rotation, the description of the resulting figure is:
"It is a square with vertices {A', B', C', D'} and with additional line segments {A' to midpoint(B',C'), B' to midpoint(C',D'), C' to midpoint(D',A'), D' to midpoint(A',B')}."
Moreover, since A, B, C, and D are vertices of a square, we know that there's a 90 degree rotation for which A' = B, B' = C, C' = D, and D' = A. Plugging these into our description of the figure, we have:
"It is a square with vertices {B, C, D, A} and with additional line segments {B to midpoint(C,D), C to midpoint(D,A), D to midpoint(A,B), A to midpoint(B,C)}."
But this is exactly what we started with, other than a reordering of the terms listed in curly braces, and this reordering makes no difference. (I deliberately chose the curly-brace notation above to suggest that these are sets, rather than ordered lists.)
So, we see that a 90-degree rotation just gives us back the description we started with, and since this description fully defines the figure, we've proven that it has the desired rotational symmetry.
