For reference: In the figure, calculate $S_x$, knowing that $DILO\space$ is a parallelogram and $S_1 + S_2 +S_3= 8m^2$ (Answer: $8m^2$)
My progress: I traced to the perpendicular $CH$ and marked the areas indicated below
Let: $S_a = S\triangle BGE\\ S_b=S\triangle LJO\\ S_c=S\triangle CEJ\\ h =LO=IO\\ S_2=\frac{BG.h}{2}\\ S_1+S_a = \frac{BG.CH}{2}\\ S_3+S_b =\frac{ HO.h}{2}\\ S_a+S_x+S_b = \frac{h.BO}{2}\\ S_c+S_x = \frac{CH.h}{2}$
I'm not able to relate the areas
As $S_2=\frac{BG.h}{2}$ and also $S_2=\frac{AB.BG}{2}$, comparing the equations we get $AB=h$
Thus we have:- $$S_3+S_b =\frac{ HO.h}{2}\\$$ $$S_a+S_x+S_b = \frac{h.BO}{2}\\$$ $$S_1+S_2+S_a=\frac{h⋅BH}{2}$$ Adding the First and Third Equations:- $$S_3+S_b+S_1+S_2+S_a=\frac{ HO.h}{2}+\frac{h⋅BH}{2}$$ $$\rightarrow 8m^2 + S_a+S_b=\frac{(h).(HO+BH)}{2}$$ $$\rightarrow 8m^2 + \frac{h.BO}{2}-S_x=\frac{h.BO}{2}$$ This follows from the 2nd equation.
Thus we get $S_x=8m^2$ as required