How to prove that the midpoints in a square generate a one twentieth triangle area? I found this identity or sort of in my book, but there isn't a proof of this. The figure is attached below.

As it can be seen it is a square where it has been traced four lines, going from all their corners to the midpoints of their opposing sides.
The letter $S$ represents I believe the surface. But I have no idea how one is three times of the other.
But again, I feel confused. How exactly the area of orange color is just one third of their contiguous trapezoid?.
Can someone explain this to me?. I tried to do it on my own and I believe the identity which comes into play is that when you have two triangles which share the same height and their bases are proportional to the other their respective areas also share the same proportion.
I appreciate that the answer can use some drawing so I can spot what's going on, because here are too many lines and I am confused. How can this be proven relying only in euclidean geometry?. Help, anyone?. Please.
 A: 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.
A: The middle square is filled with four copies of the shaded area marked $S$. You can see there why the trapezoid in the upper right has area $3S$.
