On the diagram the points that divide each of the sides into equal thirds can be marked. The lines connecting these points to the nearby vertex of the yellow triangle can also be drawn. This gives the following diagram:

Now, triangle $AHE$ is an enlargement of $ABC$ by scale factor $\tfrac{1}{3}$, as $AE=\tfrac{1}{3}AC$, $AH=\tfrac{1}{3}AB$ and $\angle EAH = \angle CAB$.

This means the area of $EAH$ is $\left( \tfrac{1}{3} \right)^2 = \tfrac{1}{9}$ of the area of $ABC$.

Since $HF = AH$, $EAH$ and $EHF$ have the same base length and the same perpendicular height (that of $E$ above $AB$), they have the same area: $\tfrac {1}{9}$ of the total area of $ABC$.

This process can be repeated at vertices $B$ and $C$, so each of the six orange triangles all have area $\tfrac{1}{9}$ of that of $ABC$.

Therefore the yellow area is $1-6\times \tfrac{1}{9} = \tfrac{1}{3}$ of the area of the whole triangle.

Steven, from Sunderland College, sent us a solution which used the sine rule for area instead. You can see his solution here.