Answer: B

Video solution:

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Written solution:

Let the square have side length $s,$ and let $H$ be the foot of the perpendicular from $F$ to $\overline{BC}.$ The right triangles $CAB$ and $CFH$ are similar, so $CH=FH.$ Also, triangles $BFH$ and $BEC$ are similar, giving $$\frac{FH}{EC}=\frac{BH}{BC}.$$ Since $EC=s/2,$ $BC=s,$ and $BH=s-CH=s-FH,$ this becomes $2FH/s=1-FH/s,$ so $FH=CH=s/3.$

Thus triangle $EFC$ has base $EC=s/2$ and height $CH=s/3,$ so its area is $s^2/12.$ Therefore $$\begin{aligned} [AFED]&=[\triangle ACD]-[\triangle EFC]\\ &=\dfrac{s^2}{2}-\dfrac{s^2}{12}\\ &=\dfrac{5s^2}{12}. \end{aligned}$$ Since this area is $45$, we get $s^2=108$, the area of the square.