MathFun4Kids

(1) Without loss of generality, assume $AB\le DE$. Select $O$ between $D$ and $E$ so that $OD=CE$. And draw line $DB$, $FF$, $BF$, $OB$, and $OF$, as shown below:

Because $OG=DG-OD=DE-CE=CD=AB$, $AO=AD+OD=DC+CE=DE=GF$, and $\angle{BAD}=\angle{FGD}=90^\circ$, $\triangle{BAO}\cong\triangle{OGF}$. Therefore, $OB=OF$.

Additionally, because $\angle{BOF}=180^\circ-\angle{AOB}-\angle{FOG}=180^\circ-(\angle{AOB}+\angle{ABO})=180^\circ-90^\circ=90^\circ$. Therefore, $\triangle{BOF}$ is an isosceles right triangle. $O$ is the center of the semi-triangle. And $OB=OF=1$, $BF=\sqrt{2}$.

Becaue $\angle{ADB}=\angle{GDF}=45^\circ$, we have $\angle{BDF}=180^\circ -(\angle{ADB}+\angle{GDF})=180^\circ-90^\circ=90^\circ$.

Therefore, $\triangle{BDF}$ is right. We have $BD^2+DF^2=BF^2=2$.

So the sum of the areas of two squares is: $$AB^2+DE^2=\dfrac{1}{2}BD^2+\dfrac{1}{2}DF^2=\dfrac{1}{2}(BD^2+DF^2)=\dfrac{1}{2}\cdot 2=\boxed{1}$$

(2) Without loss of generality, assume $AB\le DE$. In order to maximize the difference between the area of $DEFG$ and $ABCD$, $EF$ must be as long as possible, which puts both $E$ and $F$ on the semi-circle, as shown below:

Therefore, $OD=OG=\dfrac{1}{2}EF$, as $EF$ is a chord of the semi-circle. We have $DE^2+OD^2=EO^2$, i.e. $EF^2+\dfrac{1}{4}EF^2=1$. Therefore, $EF^2=\dfrac{4}{5}$.

Hence, the area of $DEFG$ is $\dfrac{4}{5}$, and that of $ABCD$ is $1-\dfrac{4}{5}=\dfrac{1}{5}$. This results in the smallest area ratio between the two squares as $\boxed{\dfrac{1}{4}}$.

Proof: Rotating $\triangle{ADC}$ counter-clock-wise around $C$ by $60^\circ$, so that $D$ is moved to $D’$, and $A$ to $B$, we have:

Because $\angle{DCD’}=60^\circ$, and $CD=CD’$, $\triangle{DCD’}$ is equilateral. Therefore, $CD=DD’$, and $\angle{CDD’}=60^\circ$.

Because $AD=BD’$, therefore the line segment $AD$, $BD$ and $CD$ forms $\triangle{DBD’}$.

Because $\angle{BDC}=150^\circ$, $\angle{CDD’}=60^\circ$, we have $$\angle{BDD’}=\angle{BDC}-\angle{CDD’}=150^\circ-60^\circ=90^\circ$$

Therefore, $\triangle{DBD’}$ is right, implying the line segment $AD$, $BD$ and $CD$ are the sides of a right triangle.