How big is the smallest triangle inscribed in a square A square is divided in 7 areas as show on the figure.
The dots show the corners of the square and the middle points on the edges.

How large a fraction is area $D$ and how do I work it out?
I have tried using trigonometry to calculate the area A. If we say each side of the square is $1$, and look at the triangle $ABC$ using Pythagoras, it's hypotenuse must be $ \sqrt {1 \cdot 1 + 0.5 \cdot 0.5} = 1.118$.
Then using the sine relation we see that the $ \hat A$ is $\sin A = 1/1.118 = 63.43°$. It then follows the other angles must be $90°$ and $71.57°$. If I use Heron's formula I can calculate the area of $A = \sqrt {p(p−a)(p−b)(p−c)} = \frac 1 {12}$.
I know $C$ is $ \frac1 {16}$ just by looking at the figure. 
The area of $ABC$ is $ \frac 1 4 $, so $B$ must be $\frac 1 4 - C - A = \frac 5 {48}$.
Now the area of $BD$ must be $ \frac 1 8$. It therefore follows that $D = \frac 1 8 - \frac 5 {48} = \frac 1 {48}$.
The trigonometry part just seems too elaborate, and I was wondering if there is a much more simple solution I am missing?

Alternatively, we can find that the triangle containing $D$ and $E$ has a base of $1$ and a height of $\frac{2}{3}$ (obtained by solving $1-x = \frac{1}{2} + \frac{1}{2}x$), so it has an area of $\frac{1}{2} \times 1 \times \frac{2}{3} = \frac{1}{3}$.
The height of $D$ is $\frac{2}{3} - \frac{1}{2} = \frac{1}{6}$, so the sides are $\frac{1}{4}$ of the big triangle. Since triangles $D$ and $D + E$ are similar by AA, the area of $D$ is $\frac{1}{16}$ times smaller. This gives the area of triangle $D$ as $\frac{1}{3} \times \frac{1}{16} = \frac{1}{48}$.
