Video solution:

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Written solution:

Let $H$ be the point on $\overline{BC}$ where the altitude from $F$ to $\overline{BC}$ meets $\overline{BC}.$ This altitude, $\overline{FH}$ is illustrated above. Then, by angle-angle similarity, we can see that $$\triangle CAB \sim \triangle CFH$$ and $$\triangle BFH \sim \triangle BEC .$$

Since the sides of similar triangles are proportional, we know that $$\dfrac{FH}{BH} = \dfrac{EC}{BC}$$and$$\dfrac{FH}{HC} = \dfrac{AB}{BC} .$$ Thus, $$\dfrac{FH}{EC} = \dfrac{BH}{BC}$$and$$\dfrac{FH}{AB} = \dfrac{HC}{BC} .$$ Adding these equations yields: $$\begin{align*}\dfrac{FH}{EC} + \dfrac{FH}{AB} &= \dfrac{BH}{BC} + \dfrac{HC}{BC}\\ &= \frac{BH+HC}{BC} \\&= \frac{BC}{BC} \\&= 1.\end{align*}$$ This, in turn, shows that $$\frac{1}{EC} + \frac{1}{AB} = \frac{1}{FH} .$$

Now, let $s$ be the side length of the square. We know $AB = 2\cdot EC = s.$ This means $$\begin{align*}\dfrac{1}{FH} &= \dfrac{1}{EC} + \frac{1}{AB} \\&= \frac{2}{s} + \frac{1}{s} \\&= \frac{3}{s} .\end{align*}$$ Therefore, $FH = \frac{s}{3} .$

Now, to compute the area of $\triangle EFC,$ we take the area of $\triangle BCE$ and subtract the area of $\triangle BFC.$ This is equal to

$$\begin{align*} \dfrac{BC\cdot EC}{2} - \dfrac{BC\cdot FH}{2} &= \dfrac{BC\cdot(EC-FH)}{2} \\ &= \dfrac{s\cdot\left(\dfrac{s}{2} - \dfrac{s}{3}\right)}{2} \\ &= \dfrac{s\cdot\frac{s}{6}}{2} \\ &= \dfrac{s^2}{12}. \end{align*}$$

The area of $AFED$ is the area of $\triangle ACD$ minus the area of $\triangle EFC,$ which is equal to$$\begin{align*} \dfrac{s^2}{2}-\dfrac{s^2}{12} &= \dfrac{5s^2}{12} \\&= 45.\end{align*}$$ With $\frac{5}{12} s^2 = 45 ,$ we get $s^2 = 108,$ which is the area of the full square.