I like Right Triangles

Since $\triangle CGH \sim \triangle CFE$ ,

$\dfrac{GH}{CH}=\dfrac{FE}{CE}$

$\dfrac{GH}{CH}=\dfrac{8}{6}$

$CH=\dfrac{3}{4}GH$

Since $\triangle DGH \sim \triangle DAB$ ,

$\dfrac{GH}{HD}=\dfrac{AB}{BD}$

$\dfrac{GH}{HD}=\dfrac{12}{6}$

$HD=\dfrac{1}{2}GH$

We know that $CH+HD=3$ , so

$\dfrac{3}{4}GH+\dfrac{1}{2}GH=3$

$GH=\dfrac{12}{5}$

The desired area is

$A=\dfrac{1}{2} (CD)(GH)=\dfrac{1}{2}(3)\left(\dfrac{5}{12}\right)=\boxed{3.6}$

Place the whole diagram on a coordinate plane such that $B$ is $(0, 0)$ . Then with the given information, $A$ is $(0, 12)$ , $C$ is $(3, 0)$ , $D$ is $(6, 0)$ , $E$ is $(9, 0)$ , and $F$ is $(9, 8)$ .

Then $CF$ is on $y = \frac{4}{3}x - 4$ and $AD$ is on $y = -2x + 12$ , and their intersection at $G$ is $(\frac{24}{5}, \frac{12}{5})$ .

Therefore, $\triangle CDG$ has a base of $3$ and a height of $\frac{12}{5}$ , and an area of $A = \frac{1}{2} \cdot 3 \cdot \frac{12}{5} = \frac{18}{5} = \boxed{3.6}$ .

Since $\triangle$ CEF ~ $\triangle$ CGH

Therefore,

$\frac{FE}{GH}$ = $\frac{CE}{CH}$

=> $\frac{8}{GH}$ = $\frac{6}{CH}$

=> 8(CH) = 6(GH) (1)

Since $\triangle$ DAB ~ $\triangle$ DGH

Therefore, $\frac{AB}{GH}$ = $\frac{BD}{HD}$

=> $\frac{AB}{GH}$ = $\frac{BD}{CD-CH}$

=> $\frac{12}{GH}$ = $\frac{6}{3- CH}$

=> 12(3-CH) = 6(GH)

=> 36-12(CH) = 6(GH) (2)

Combine (1) and (2) through Transitive Property, we got:

36-12(CH) = 8(CH)

=>36= 20(CH)

=>CH = 1.8

We have that: 8(CH) = 6(GH)

=>8(1.8) = 6(GH)

=>GH= 2.4

Area of $\triangle$ CGD = $\frac{CD*GH}{2}$ = $\frac{3 \times 2.4}{2}$ = $\boxed{3.6}$