A special triangle corrected

$CF=DC=\frac12$

Let side of small square be $IH=a$

Ratio $\frac{ED}{BD}=1$ , therefore $\frac{IH}{HE}=1$ and $HE=a$

$HF=FG=\frac a2$

$EF=EH+HF=a+\frac a2=\frac{3a}2$

But we know that $EF=\frac12$ , that gives us the equation for $a$ as: $\frac{3a}2=\frac12$ and $a=\frac13$

$FG=\frac a2=\frac16$

Area $[CFG]=\frac12\times CF\times FG=\frac12\times\frac12\times\frac16=\frac1{24}$

Inverse of the area is: $\boxed{24}$

Considering that $BC=1$ , with $D$ as the midpoint of $BC$ , we have $BD=CD=\frac{1}{2}$ . Then $CDEF$ being a square means $DE=EF=CF=CD=\frac{1}{2}$ . Let $FG=x$ . Then $HF=FG=x$ as given in the question which means $EH=EF-HF=\frac{1}{2}-x$ .

Now note that $BDE$ and $EHI$ are similar triangles as $BD$ and $EH$ are parallel and so are $DE$ and $HI$ . Therefore $\frac{EH}{HI}=\frac{BD}{DE}=1$ which means $HI=EH=\frac{1}{2}-x$ . Considering $GHIJ$ is also a square, $GH=HI=\frac{1}{2}-x$ . Moreover, $GH=HF+FG=x+x=(1+1)x=2x$ . Therefore:

$2x=\frac{1}{2}-x$

$2x+x=\frac{1}{2}$

$(2+1)x=\frac{1}{2}$

$3x=\frac{1}{2}$

$x=\frac{1}{3\times2}$

$x=\frac{1}{6}$

Therefore, the area of $CFG=\frac{1}{2} \times CF \times FG=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{6}$ , the reciprocal of which is $2\times 2\times 6=4\times 6=\boxed{24}$