Triangle of Intersections

Through successive shear mappings , $\triangle ABC$ can be transformed to unit equilateral triangle $\triangle A'B'C'$ while still preserving all of its area and length ratios.

Since $\frac{BD}{BC} = \frac{CE}{AC} = \frac{AF}{AB}$ , the diagram above has rotational symmetry, so $\triangle R'Q'S'$ is an equilateral triangle, $B'D' = C'E'$ , and $D'R' = E'Q'$ .

Since $\triangle R'Q'S'$ and $\triangle A'B'C'$ are both equilateral triangles, and since $\triangle RQS$ is half the area of $\triangle ABC$ , and since the sides of $\triangle A'B'C'$ are $1$ , the sides of $\triangle R'Q'S'$ must be $\frac{\sqrt{2}}{2}$ .

Also since $\triangle R'Q'S'$ and $\triangle A'B'C'$ are both equilateral triangles, $\angle A'C'B' = 60°$ and $\angle Q'R'S' = 60°$ , and by vertical angles $\angle B'R'D' = 60°$ as well. This means that $\triangle B'R'D' \sim \triangle B'C'E'$ by AA similarity, so $\frac{D'R'}{B'R'} = \frac{C'E'}{B'C'}$ and $\frac{B'D'}{B'R'} = \frac{B'E'}{B'C'}$ . Substituting $C'E' = B'D'$ , $B'C' = 1$ , and $B'E' = B'R' + \frac{\sqrt{2}}{2} + D'R'$ gives $\frac{D'R'}{B'R'} = B'D'$ and $\frac{B'D'}{B'R'} = B'R' + \frac{\sqrt{2}}{2} + D'R'$ . Furthermore, using the law of the cosines on $\triangle B'R'D$ gives $B'D'^2 = B'R'^2 + D'R'^2 - B'R' \cdot D'R'$ .

These three equations solve to $B'D' = \alpha = \frac{21 - 3\sqrt{21}}{42} \approx \boxed{0.173}$ .