Find the area of ​the $AMND$ quadrilateral;

Question

For refrence: In the ABCD trapeze, (BC < AD) it is known that the area of the $BO$C triangle is $9 m^2$ and the area of ​​the triangle $AOD=16m^2$ ($O$ is the intersection point of the diagonals). Calculate the area of ​​the $AMND$ quadrilateral; if $AM = ME(M \in AE$); $EN = ND(N \in ED)$ and $AB\cap DC = {E}$. (Answer:$84m^2$)

My progress:

$S_{AOB} = S_{COD}\\ S_{BOC}.S_{AOD}=S_{AOB}^2\implies 144 = S_{AOB}^2 \\ \therefore S_{AOB}=S_{COD} =12 \implies S_{ABCD}=49\\ \frac{S_{EMN}}{S_{AED}}=\frac{EM.EN}{EA.ED}=\frac{1}{4}\\ S_{AMND} = S_{AED} - S_{EMN} = \frac{3S_{AED}}{4}$

I didn't find the relationship to calculate the $AED$ area...

Answer 1

$\triangle AOD \sim \triangle COBC$. So $BC:AD = 3:4$, as area of $\triangle AOD$ and $\triangle BOC$ are in ratio $16:9$. If altitude from $O$ to $AD$ is $h$, altitude from $O$ to $BC$ is $\frac{3h}{4}$. If altitude from $E$ to $AD$ is $h_1$, the altitude to $BC$ is $(h_1 - \frac{7h}{4})$ and

$ \displaystyle \frac{h_1 - 7h/4}{h_1} = \frac 3 4 \implies h_1 = 7 h$

As base of both triangle $\triangle AOD$ and $\triangle EAD$ is same, $S_{\triangle EAD} = 7 \cdot S_{\triangle AOD} = 16 \cdot 7$

Now use the relationship between area of $AMND$ and $\triangle EAD$ you obtained.

Answer 2

Hint :

Establish that $MN:BC:AD=2:3:4$ . Then from similar triangles, it follows that $$[EMN] : [EBC] : [EAD] = 2^2:3^2:4^2$$ $$\Rightarrow \frac{[AMND]}{[ABCD]}=\frac{4^2-2^2}{4^2-3^2}$$