Appearances are sometimes deceiving

Segments $DE,EF,DF$ divide $\triangle ABC$ into four congruent triangles. Points $G,H,I$ area orthocentres of triangles $EFA, EDC, DFB$ respectively. Hexagon $EHDIFG$ can be divided into four triangles: $DEF, EFG, EDH, DFI$ . Area of $\triangle DEF$ is $\frac{1}{4}$ of the area of $\triangle ABC$ . Triangles $EFG, EDH, DFI$ make up the area of whole congruent triangle i.e. $EFA$ as their vertices are made up of orthocentre and each pair of the reference triangle sides. This means that hexagon area is equal to areas of two smaller triangles like that of $DEF$ and $EFA$ which in turn is half of the area of triangle $ABC$ . The answer is $\boxed{\frac{1}{2}}$ .

Note:

1) All vertices of the hexagon lie on the nine-point circle of the large triangle.

2) The ratio is the same regardless of the angle measures.