jmc

It can be proven that $\Delta ADX\cong \Delta JIX$ (where $X$ is the point where $\overline{AJ}$ intersects $\overline{HC}$) which also means quadrilaterals $ABCX\cong JGHX$ (due to the squares being equal in the area which means the squares are congruent, and since the triangles earlier mentioned are congruent). The area of the shaded region is equal to the area of one square since the quadrilaterals and triangles are congruent. The total area of the shape of three squares. Putting these two pieces of information together, the answer is $\boxed{\frac{1}{3}}$.