Find the area of ​the CEN triangular region

Question

For reference: If ABCD is a square, AB = DE. Calculate the area of ​​the CEN triangular region.(Answer:$6m^2$)

My progress:

$S_{\triangle MCE}=X=\frac{MC⋅DC}{2}=\frac{MC⋅ℓ}{2}(I)\\ I+S_{\triangle MBA}=\frac{l^2}{2}\\ \therefore X+5 = \frac{l^2}{2}\implies 9 + S_{\triangle CEN} = \frac{l^2}{2}\\ \frac{4}{MN}=\frac{S_{\triangleÇNE}}{EN}=\frac{CN.DE}{2EN}\\ \frac{5}{BM}=\frac{4}{MC}\\ S_{ABCE} = \frac{3AB^2}{2}$

...?

Answer 1

Set: $x=BM$, $y=MC$, $s=S_{CNE}$ (in m$^2$). We have: $$ x:y=5:(4+s),\quad (x+y):y=s:4. $$ Combine these to get $5:(4+s)=(s-4):4$.

Answer 2

You can use a system of $4$ equations :

$\begin{cases} \frac{l \cdot BM}{2} = 5 \\ \frac{(l -BM)\cdot CN}{2} = 4 \\ \frac{(l -BM)\cdot l}{2} = A_{CNE}+4 \\ \frac{l\cdot CN}{2} = A_{CNE} \end{cases} $

and solve for $A_{CNE}$.

Answer 3

Draw $MD$

$S_{MND} = A\\ S_{DEN}=A_x\\ AD(median)\implies {S_{AMD}} = A+ A_X\\ MDEC\implies A = S_{CNE}(by~ property)\\ S_{AMD} = \frac{l^2}{2}\implies S_{ABM}+S_{MCD}=S_{AMD} \\ \therefore A+A_X = 5+4+A\implies A_X =9\\ \therefore MDEC: A_X.4 = A^2(by~ property) \implies \boxed{\color{red}A = 6m^2} $