Find the ratio between the areas of $ABCE$ and $CEN $

Question

In a pentagon $ABCDE$, extension $BC$ and $ED$ intersect at $N$. Calculate the ratio between the areas of $ABCE$ and $CEN$.($S:1$)

I try:

$S_\triangle CEN =\dfrac{CE\cdot h}{2}(I)$

$ S_{ABCE} =\dfrac{AB+CE}{2}\cdot h_1 (II)$

$\dfrac{II}{I} = \dfrac{(AB+CE)\cdot h_1}{CE\cdot h}$

$ \triangle CEN_{(isosceles)}: \implies CN =CE$

$ \cos(36^{\circ}) = \dfrac{EB}{2AE}\implies \dfrac{\sqrt{5}+1}{4} = \dfrac{EB}{2AE} \therefore EB=\dfrac{AE(\sqrt5+1)}{2}$

Answer 1

$\begin{cases}EB=EC\\AB=AE=ED=DC\end{cases}$$\Rightarrow \triangle ABC=\triangle EDC$ for SSS.

$\begin{cases}DC=BC\\\angle NDC=ECB=72^o\\\angle NCD=108-36=72^o=\angle ECB\end{cases}$$\Rightarrow \triangle NDC=\triangle EBC$ for ASA

therefore:

$$\frac{S_{ABCE}}{S_{NCE}}=1$$

Answer 2

$$ABCE=ABCDE-CED=ABCDE-ABE$$and $$CEN=CBED=ABCDE-ABE$$Therefore$$\frac{ABCE}{CEN}=\frac{1}{1}$$