Similar Triangles Might Help?

Since $\overline{AE}$ is parallel to $\overline{BC},$ $\angle FBC = \angle FED \text{ and } \angle FCB = \angle FDE=90^\circ,$ implying $\triangle FBC$ and $\triangle FED$ are similar triangles. Also, since $\lvert \overline{BC}\rvert = \lvert \overline{CD}\rvert$ , we have $\lvert \overline{DF}\rvert = \lvert \overline{BC}\rvert - \lvert \overline{CF}\rvert = 30 - 20 = 10.$ Hence, $\begin{aligned} \lvert \overline{CF}\rvert : \lvert \overline{DF}\rvert &= \lvert \overline{BC}\rvert : \lvert \overline{DE}\rvert \\ 20 : 10 &= 30 : \lvert \overline{DE}\rvert \\ \lvert \overline{DE}\rvert &= 15. \end{aligned}$

Let $\lvert\triangle CEF\rvert$ denote the area of $\triangle CEF,$ and $\lvert\square ABCE\rvert$ the area of $\square ABCE.$ Then since $\lvert\triangle CEF\rvert=\lvert\square ABCE\rvert - (\lvert\square ABCD\rvert + \lvert\triangle DEF\rvert),$ we have $\begin{aligned} \text{Area of }\triangle CEF &= \lvert\square ABCE\rvert - (\lvert\square ABCD\rvert + \lvert\triangle DEF\rvert) \\ &= \frac{1}{2} \times \lvert \overline{AB}\rvert \times \left(\lvert \overline{BC}\rvert + \lvert \overline{AE}\rvert\right)- \left(\lvert \overline{BC}\rvert^2 + \frac{1}{2} \times \lvert \overline{DE}\rvert \times \lvert \overline{DF}\rvert\right) \\ &= \frac{1}{2} \times 30 \times (30 + 45) - (900 + 75) \\ &= 150. \end{aligned}$

Since $\triangle BFC \sim \triangle DFE$ , we have

$\dfrac{x}{10}=\dfrac{30}{20}$ $\implies$ $x=\dfrac{300}{20}=15$

$A_{CEF}=45(30)-\dfrac{1}{2}(45)(30)-\dfrac{1}{2}(30)(20)-\dfrac{1}{2}(30)(15)=1350-675-300-225=$ $\boxed{150}$

Note:

1. The area of $\triangle CEF$ is equal to the area of rectangle $AEGB$ minus the area of $\triangle BAE$ minus the area of $\triangle BFC$ minus the area of $\triangle CEG$ .

A(CEF) = A(BEC) - A(BFC) = 0.5(BC)(CD) - 0.5(BC)(FC) = 0.5(30)(30) - 0.5(30)(20) = 150

From the given information, we know that FD = 10, because CD = 30 and CF = 20 (subtraction).

We know that triangles FDE and BAE are similar, because both share angle DEF and are right triangles, and and DE and AB are parallel, so angles DFE and ABE are equal).

Therefore side 1/side 2 of triangle ABE = side 1/side 2 of triangle FDE, and let DE = x.
30/(30+x) = 10/x --> 30x = 300 + 10x --> 20x = 300 --> x = 15

By the triangle formula, 1/2bh, the area of CFE is (1/2)(20)(15) = 150.

Notice that $\angle AEB = \angle EBC$ , so $\frac{FC}{BC} = \frac{AB}{AE} \implies \frac{20}{30} = \frac{30}{AE}$ or $AE = 45$

Solving for the area of $A_{CEF}$ ,

$A_{CEF}=A_{AECB} - A_{AEB} - A_{BCF}=[\frac{1}{2}(30 + 45)(30)] - \frac{1}{2}(30)(45) - \frac{1}{2}(20)(30)=$ $\boxed{150}$