jmc

Let $u=BE$, $v=CE$, $x=CF$, and $y=DF$. Then the area of triangle $ABE$ is $u(x+y)/2=8$, so $u(x+y)=16$. The area of triangle $ADF$ is $y(u+v)/2=5$, so $y(u+v)=10$. The area of triangle $CEF$ is $xv/2=9$, so $xv=18$. Thus, we have the system of equations $$\begin{array}{rl}ux+uy& =16,\\ uy+vy& =10,\\ vx& =18.\end{array}$$ Solving for $x$ in equation (1), we find $$x=\frac{16-uy}{u}.$$ Solving for $v$ in equation (2), we find $$v=\frac{10-uy}{y}.$$ Substituting into equation (3), we get $$\frac{10-uy}{y}\cdot \frac{16-uy}{u}=18.$$ This equation simplifies to $$u^2{y}^2-44uy+160=0.$$ We recognize this equation as a quadratic in $uy$, which factors as $(uy-4)(uy-40)=0$. From equation (1), $uy$ must be less than 16, so $uy=4$.

Then from equation (1), $ux=16-uy=16-4=12$, and from equation (2), $vy=10-uy=10-4=6$. Therefore, the area of rectangle $ABCD$ is $(u+v)(x+y)=ux+uy+vx+vy=12+4+18+6=\boxed{40}$.