$ABB_1A_1,BB_2C_1C,ACC_2A_2$ are squares. The problem itself is to prove that the area of $ABC$ and the area of $BB_1B_2,CC_1C_2,AA_1A_2$ are equal. If I could only prove that GEDB is a square it would follow that $\triangle AEB \cong \triangle DBB_1$ and thus the area of $\triangle ABC$ would equal that of $\triangle BB_1B_2 $ and I guess the case with the other triangles would follow similarly. Any Ideas on how to prove this?
EDIT: You can only use the formulas for the area of a square $A = a^2$ and that of a triangle $A = \frac{a\times h_a}{2}$ and of course all of the properties of triangles,squares and the congruence theorems\postulates.
How about this: extend $AB$, $BB_1$, and $EB$ so as to show $\angle{DBB_1}=\angle{ABE}$ by ASA using the fact that we know that $\angle{ABB_1}=\angle{CBB_2}$. Then $\bigtriangleup DBB_1 \cong \bigtriangleup AEB$ and therefore $DB=EB$. And $DB \perp EB$ by definition so we're done.