Let $F,G$ are on $AC$ and $BF \perp AC,DG \perp AC$. Then $\frac{BE}{ED}=\frac{BF}{DG}=\frac{S(ABC)}{S(ADC)}=\frac{AB*BC* \sin{\angle ABC}}{AD*CD*\sin{\angle ADC}}=\frac{BE}{ED} *\frac{ \sin{\angle ABC}}{\sin{\angle ADC}} \to \sin{\angle ABC}=\sin{\angle ADC}$. So $\angle ABC = \angle ADC$ or $\angle ABC+\angle ADC=180$

RagvaloD wrote: Let $F,G$ are on $AC$ and $BF \perp AC,DG \perp AC$. Then $\frac{BE}{ED}=\frac{BF}{DG}=\frac{S(ABC)}{S(ADC)}=\frac{AB*BC* \sin{\angle ABC}}{AD*CD*\sin{\angle ADC}}=\frac{BE}{ED} *\frac{ \sin{\angle ABC}}{\sin{\angle ADC}} \to \sin{\angle ABC}=\sin{\angle ADC}$. So $\angle ABC = \angle ADC$ or $\angle ABC+\angle ADC=180$ It is perfect, thank you

Let the common ratio be $k$ so that $AB=ka, BC=kb, CD=a, DA=b, BE=k^2c,$ and $ED=c$. Stuart's Theorem on $\triangle BCD$ and $\triangle ABD$ give $BD(CE^2+k^2c^2)=k^2b^2c+k^2a^2c$ and $BD(AE^2+k^2c^2)=k^2b^2c+k^2a^2c$. Subtracting the latter from the prior gives $CE=AE.$ Area Lemma: Let $ABC$ be a triangle where $D\in BC$ and $X\in AD.$ Then, $\frac{[ABC]}{[BCX]}=\frac{AD}{DX}.$ Proof: Let $A'$ be the foot from $A$ to line $BC$ and let $X'$ be the foot from $X$ to line $BC$. Then, $\frac{[ABC]}{[BCX]}=\frac{\frac{1}{2}\cdot AA'\cdot BC}{\frac{1}{2}\cdot XX'\cdot BC}=\frac{AA'}{XX'}=\frac{AD}{DX}$, as desired. $\Box$ Hence, by the Area Lemma and the fact that $CE=AE$ gives that $[CBD]=[BAD]$. But, $[CBD]=\frac{1}{2}\cdot kb\cdot a\cdot \sin C$ and $[BAD]=\frac{1}{2}\cdot ka\cdot b\cdot \sin A$. Thus, $\sin A=\sin C$ so either $ABCD$ is a cyclic quadrilateral or parallelogram, as desired. $\blacksquare$ Remark: Note that $ABCD$ can be a parallelogram if and only if $k=1$.

I hope the solution is correct.

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