$ \frac {AD} {DC}=\frac {BD} {DE} \Rightarrow \triangle ADC \sim \triangle BDE \Rightarrow \angle DCA=\angle DEB$ and $ \angle DAC=\angle DBE$ Denote by $ P=BE \cap AC$ $ \angle ACB=\angle ABP \Rightarrow \triangle APB \sim \triangle ABC \Rightarrow \angle ABC=\angle APB=\angle ADB$ In the quadrileteral $ ABCE$ the sum of the angles is equal with $ 180^{\circ}$ $ (\angle CEA+\angle EAB+\angle ABC)+\angle BCE=360^{\circ} \Rightarrow 180^{\circ}+\angle BCE=360^{\circ} \Rightarrow$ $ \Rightarrow \angle BCE=180^{\circ}$

moldovan wrote: In the quadrileteral $ ABCE$ the sum of the angles is equal with $ 180^{\circ}$ $ (\angle CEA+\angle EAB+\angle ABC)+\angle BCE=360^{\circ} \Rightarrow 180^{\circ}+\angle BCE=360^{\circ} \Rightarrow$ $ \Rightarrow \angle BCE=180^{\circ}$ Can you explain the remark above for me ?

The second angle equality was sufficient for the requirement of the problem, that makes $ABCD$ cyclic, done. Best regards, sunken rock

sunken rock wrote: The second angle equality was sufficient for the requirement of the problem, that makes $ABCD$ cyclic, done. Best regards, sunken rock What is the angle equality which you talk about? Can you write it?

Pls, disregard my previous post, I was not ON! Best regards, sunken rock

Let $AC\cap BE=X, AB\cap CE=Y$. By first two angle conditions $\triangle DAB\sim \triangle DCE$, means $D$ is Miquel point of $ABEC$. So $ECXD, ACYD$ are cyclic. Since $D\in AE$, by Miquel on $\triangle ABE$ we have $BXCY$ - also cyclic. So $\angle ACY = \angle XBY=\angle EBA=\angle ACB$ , which is what we need.