The diagram shows the red and green triangles and the \((x, y)\) coordinates of the points \(A\), \(B\), \(C\), \(D\), \(E\) and \(F\). Point B is at position \((b, h)\), point C at \((c, 0)\) and the points D, E and F divide the three sides in the ratios of \(r:s\) with \(r + s = 1\) (in our case \(r=1/3\) and \(s=2/3\)).

We can find the coordinates of point G from the intersection of the vectors CD and AE:\[(c,0)+k(rb-c,rh)=\overline{k}(sb+rc, sh)\]where \(k\) and \(\overline{k}\) are determined by equating the \(x\) and \(y\) coordinates, which gives the coordinates of point G as:\[G=\frac{r}{1-rs}\left(sb+rc, sh\right)\]In the same way, the coordinates of point H are determined from the intersection of the vectors FB and AE as:\[H=\frac{s}{1-rs}\left(sb+rc, sh\right)\]The intersection of the vectors CD and FB gives point I’s coordinates as:\[I=\frac{1}{1-rs}(r^2b+s^2c,r^2h)\]

We now have the heights of the points G, H and I above the baseline AC so we can calculate the areas of triangles ACG, AFH and FCI. If we let the symbol \(\bigtriangleup\) stand for the area of a triangle divided by the area of the red triangle ABC, we can now show that:\[\bigtriangleup_{ACG}=\frac{rs}{1-rs}\] \[\bigtriangleup_{AFH}=\frac{s^3}{1-rs}\]\[\bigtriangleup_{FCI}=\frac{r^3}{1-rs}\] We can now calculate the area of the green triangle \(\bigtriangleup_{AFH}-\bigtriangleup_{ACG}+\bigtriangleup_{FCI}\) as (noting that \(r^2+s \equiv s^2+r\equiv 1-rs\)):\[\frac{1-4rs}{1-rs}\] Setting \(r=1/3\) and \(s=2/3\) shows that the area of the green triangle is one seventh of that of the red triangle.