What is the relationship between blue area and red area? This is a grade school problem!
Consider the following figure:

It is very easy to show that the red area and the blue area equal. I can demonstrate this based on my knowledge related to the computation of the surface areas of circular sectors and triangles. Both areas equal $2\left(\frac{r^2\pi}4-\frac{r^2}2\right)$ where $r$ is the radius of the smaller circles.


But, how am I going to show the same if I forget, for good, the formula providing the area of a triangle?

I am not able to get rid of my thought process using triangles.
 A: See the image with blue parts shifted:


The single blue figure is the same part of its small square as the red plus both blue of the big square, hence areas
$$\frac{2\cdot blue + red}{blue}=\frac{big\ square}{small\ square}=4$$
so
$$2\cdot blue + red = 4\cdot blue$$
hence
$$red = 2\cdot blue$$
Q.E.D.
Without the formula for the area of a triangle, and without the formula for the area of a circle...
A: If you forgot formulas, but remember that areas are quadratic (i.e. "double the length, quadruple the area"), then you know that a circle of radius $2r$ uses four times the area of circle of radius $r$. Here, you have one quarter of a circle of radius $2r$, and one circle of radius $r$ (in two halves), so they should cover the same area. In other words, the two halves of the small circle should be enough to cover the large quarter circle.
However, the two halves of the small circle overlap (the blue area) and thus some of the area covering power is lost. And, indeed, the big quarter-circle is partially uncovered (the red area). Therefore, the blue and red area MUST be equal.
A: The red area $R$ is the area of the big quarter circle minus the internal white and blue areas. The radius of the larger circle is 2*r. So the area of the big quarter circle is:
$$ Q = \frac {\pi(2r)^2}{4} = \pi r^2 $$
Visually, this is equivalent to a quarter of each small circle (together, these are half a small circle) plus $r^2$ (the area of the small square in the lower-left corner) plus the red area:
$$ Q = \pi r^2 = \frac {\pi r^2}{2} + r^2 + R $$
Therefore:
$$ R = \frac {\pi r^2}{2} - r^2 $$

The blue area is equal to the area of the small square $r^2$ minus 2 times a smaller unknown white area, which we will call $S$. $S$ is equal to the area of the small square minus a quarter of the small circle.
$$ B = r^2 - 2S $$
$$ S = r^2 - \frac {\pi r^2}{4} $$
Simplify:
$$ B = r^2 - 2 \cdot (r^2 - \frac {\pi r^2}{4}) $$
$$ B = r^2 - 2r^2 + \frac {\pi r^2}{2} $$
$$ B = \frac {\pi r^2}{2} - r^2 $$
A: I assume the sides of the square are $2R$, and that the blue area is the area of the intersection of two half circles of radius $R$
$RedArea = Sa_{Big_{quartercircle}} - 2 * Sa_{Small_{halfcircles}} + intersection $
But 
$Sa_{Big_{quartercircle}} = \pi R^2$
and:
$Sa_{Small_{halfcircle}} = \pi R^2\div 2$
So:
$RedArea = intersection = BlueArea$
A: There are two red parts and two blue parts. Let one blue part be called $B$ and red be $R$. Radius of small circle be $r$. Now write area of big quarter circle with radius $2r$ in terms of these variables using inclusion exclusion principle:
$$2 \cdot \frac{\pi r^2}{2} -2B+2R = \frac{\pi (2r)^2}{4}\\ R=B$$
Note: We actually do not need that area of a disk is $\pi r^2$. We can assume it to be $A$ and use similarity of figures.
$$2 \cdot \frac{A}{2} -2B+2R = \frac{2^2A}{4} \\ R=B$$
A: Apply the carpet theorem, see https://www.cut-the-knot.org/Curriculum/Geometry/CarpetsInSquare.shtml.  Consider the upper eighth of the big circle and the left halve circle.  Since their areas are equal, the areas which are not covered by both must be equal as well, that is the upper half of the red area equals the lower half of the blue one.

A: Let r = the radius of the large quarter-circle = the side of the square. \
Area of the big quarter-circle = (\pi * r^2) / 4 \
Area of two smaller semi-circles = (\pi * r^2) / 4 \
These two areas are equal. \
Graphically, the area of the big quarter-circle (\pi * r^2) / 4 = area of two smaller semi-circles (\pi * r^2) / 4 + red shaded area (because it is contained outside the smaller circles) - blue area (because it is contained by the smaller circles twice). \
Cancel the circle areas on each side.  0 = blue area - red area \
Use addition to move the red area to the left side.  red area = blue area
