Geometry Problem

$\triangle s~AFD~and~ DEB~are ~similar~\because~of \parallel~lines.\\Sides~of~similar~\triangle s ~are~proportional ~ to~\sqrt{their~areas.}\\\therefore~AD:DB::4:3.~~~~~~So~~AD:AB::4:7\\\implies~areas~ADF:ABC::4^2:7^2~implies~area~ABC=49\\Areas~CFDE=ABC-ADF-DBE=49-16-9=24\\\boxed{ \Large 24}$

Firstly, denote CF = ED = ${y}$ , DF = CE = ${x}$ and $\angle$ CFD = $\theta$ . Now note that in terms of scale factor: $\large \Delta \ AFD = \frac{4}{3} \Delta \ DBE$ $\Rightarrow$ $\large y = \frac{3}{4} AF \ and \ x = \frac{4}{3} EB$ . This means that if we draw line CD then area [CFDE] = $\ xysin\theta$ (1). Now finding the area of $\Delta$ DBE (using $\frac{1}{2}$ AB sin $\theta$ = $\ A_{T}$ again ): $9 = \frac{3}{8}(EB).(AF)sin(180 - \theta ) = \frac{3}{8}\times\frac{3}{4}x\times\frac{4}{3}y sin\theta = \frac{3}{8} xysin\theta$ $\therefore\ (from \ (1)) \ area(CFED) = 9\times\frac{8}{3} = \boxed{24}$

CF // DE, and FD//CE, then FDEF is a parallelogram. The ratio of AD to DE is 4:3.Let the area of [CDF] =x , [CDE]=x.So the area ratio of [ADC]/[CDB] =(x+16)/(x+9) = 4/3, then x=12. So the area of CFDE = 2(12) = 24.

My approach is same of Niranjan Khanderia, ..people in same age range frequently have coincident views.