Geometry Challenge – 8 ⭐

Solution: Draw line $PQ$ passing thru $G$ and parallel to $AD$, intersecting $AB$ and $CD$ at $P$ and $Q$ respectively. Since $ABCD$ is a unit square, the total area of $\triangle{ABG}$ and $\triangle{EFG}$ is $$S=\dfrac{1}{2}(AB\cdot GP+CD\cdot DE)=\dfrac{1}{2}(GP+CD\cdot(PQ-GP))$$ $$=\dfrac{1}{2}(GP+CD\cdot(1-GP))$$

Because $\triangle{ABG}\sim\triangle{EFG}$, we have $$\dfrac{AB}{GP}=\dfrac{CD}{GQ}$$ Therefore, $$CD=\dfrac{AB\cdot GQ}{CP}=\dfrac{GQ}{GP}=\dfrac{1-GP}{GP}$$

We have $$S=\dfrac{1}{2}(GP+\dfrac{(1-GP)^2}{GP})=GP+\dfrac{1}{2\cdot GP}-1$$ $$\ge 2\cdot\sqrt{GP\cdot \dfrac{1}{2\cdot GP}}-1=\sqrt{2}-1$$

Therefore the minimum value of $S$ is $\boxed{\sqrt{2}-1}$, when $GP=\dfrac{1}{2\cdot GP}$, i.e. $GP=\dfrac{\sqrt{2}}{2}$, and $CD=\sqrt{2}-1$.