Area of a quadrilateral inside a square In the image, the segments inside the square go from a vertex to the middle point of the opposite side. If the length of the sides of the square is $1$, the area of $ABCD$ is?

Any hints?
I tried some things matching the bigger triangles but i can't get the area of $ABCD$.

If we move the figure to a $xy$ axis, we can solve this problem by finding the line equations and intersection points. Consider this image:

Since we know the point coordinates in the image, we can easily find the line equation of lines 1, 2 and 3 (we can substitute the point values in the slope-intercept form equation $y = ax + b$ and find the equation of each line).
For line 1, the line equation is $y = 2x$.
For line 2, the line equation is $y = \frac{2 - x}{2}$
For line 3, the line equation is $y = \frac{1 - x}{2}$
To find point A, we have to find the intersection point of lines 1 and 3:
$$
2x = \cfrac{1 - x}{2} \Rightarrow x = \cfrac{1}{5} \;and \; y = \cfrac{2}{5} \Rightarrow A = \left(\cfrac{1}{5}, \cfrac{2}{5}\right)
$$
We can do the same with lines 1 and 2 in order to find point B:
$$
2x = \cfrac{2 - x}{2} \Rightarrow x = \cfrac{2}{5} \; and \; y = \cfrac{4}{5} \Rightarrow B = \left(\cfrac{2}{5}, \cfrac{4}{5}\right)
$$
The side of the square (we can call it $L$) will then be the distance $\overline{AB}$:
$$
L = \overline{AB} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 } = \sqrt{\cfrac{1}{25} + \cfrac{4}{25}} = \sqrt{\cfrac{1}{5}}
$$
Finally, the area of the square will be:
$$
Area = L^2 = \cfrac{1}{5}
$$
