Trapezoid and areas

Label the figure as follows. $\triangle NKE$ and $\triangle NEM$ share a common altitude, $NF$ , thus, the ratio of their areas equals the ratio of the corresponding bases: $\frac{B}{A}=\frac{KE}{EM}\Rightarrow \frac{KE}{EM}=3$ $\triangle EKL$ and $\triangle EMN$ are similar (alternating angles), hence the ratio of their areas equals the square of their similarity ratio, i.e. $\dfrac{D}{A}={{\left( \dfrac{KE}{EM} \right)}^{2}}$ Combining, $\dfrac{D}{A}={{3}^{2}}\Rightarrow \boxed{D=9A}$