Para

There are 12 equal triangles. Four of them make the blue area, which is 4/12, which is 1/3.

Let $a,$ $b,$ $c,$ $d,$ $e,$ and $f$ be the six areas, labeled below.

The diagonal splits the parallelogram into two halves, so

$\begin{aligned} a+b+c &= \frac{1}{2} \\ d+e+f &= \frac{1}{2} \end{aligned}$

We also have the smaller triangles formed from areas at the left and on top,

$\begin{aligned} a+d &= \frac{1}{4} \\ c+f &= \frac{1}{4} \\ \end{aligned}$

Now consider that the triangle formed from area $d$ and the triangle formed from areas $b$ and $c$ are similar. Likewise, the triangle formed from area $f$ and the triangle formed from areas $a$ and $b$ are similar. This gives:

$\begin{aligned} d &= \frac{1}{4}(b+c) \\ f &= \frac{1}{4}(a+b) \\ \end{aligned}$

Now we have six distinct equations for the six unknowns. Solving the system gives

$\begin{aligned} a &= \frac{1}{6} \\ b &= \frac{1}{6} \\ c &= \frac{1}{6} \\ d &= \frac{1}{12} \\ e &= \frac{1}{3} \\ f &= \frac{1}{12} \end{aligned}$

The total blue area is $\frac{1}{6}+\frac{1}{12}+\frac{1}{12}=\boxed{\frac{1}{3}}.$