Geometry without using trigonometry

Look at the attached picture. We have that the triangles $ABC$ and $ADE$ are similar, so since $AE = 2AD$ then $AC = 2AB$. Now since there are so many right angles, all the segments marked with a single tick have the same length, and the ones marked with two ticks have twice that length. It follows that, if the big square has side lenght equal to $1$, then $AB = 1/6$. Now we can split the blue figure in four triangles, whose areas are easy to calculate: the green ones have area $1/9$, the pink ones $1/18$. Putting all those together, the area of the blue figure is \[ 2 \cdot \frac{1}{ 9} + 2 \cdot \frac{1}{18} = \frac{3}{9} = \frac{1}{3}. \]