WakeUp wrote: Let $ABCD$ be a convex quadrilateral such that $\angle ADB=\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality \[AE\cdot ED + BE^2=CD\cdot AE.\] Show that $\angle EBA=\angle DCB$. Let $F$ on the side $CD$ such that $DE=DF$ then we have $ \triangle{DEB} = \triangle{DFB} $ Therefore, $BE=BF$ and $\angle{BEA} = \angle{BFC} $ It is sufficient to show that $ \triangle{AEB} \sim \triangle{BFC} $ indeed, $ AE\cdot ED + BE^2=CD\cdot AE \Leftrightarrow AE(CD-DE) = BE.BF \Leftrightarrow AE(CD-DF) = BE.BF \Leftrightarrow AE.FC = BE.BF \Leftrightarrow \frac{AE}{EB} = \frac{BF}{FC} $ , as required.

Solved with @blastoor Let $BE$ intersect $\odot(ABD)$ at point $X \neq B$. From PoP we have that $AE \cdot ED = BE \cdot EX$. Thus original equality becomes: \begin{align*} AE \cdot ED + BE^2 = CD \cdot AE \\ BE \cdot EX + BE^2 = CD \cdot AE \\ BE \cdot BX = CD \cdot AE \qquad (2) \\ \end{align*} Claim: $\triangle BXA \sim \triangle CDB$ Proof: It is easy to see that $\triangle AEX \sim \triangle BED $ and that $\angle AXB =\angle ABD = \angle BDC$. Consequently we have: $$ \frac{AE}{AX} = \frac{BE}{BD} \implies BE \cdot AX = AE \cdot BD \qquad (20 $$Dividing relations $(1)$ and $(2)$ we get that: $$ \frac{BX}{AX} = \frac{CD}{BD} $$This implies that $\triangle BXA \sim \triangle CDB$ and thus the claim is proved. From claim follows $\angle XBA =\angle EBA =\angle DCB$ as desired.

Construct $L$ on $AD$ such that $DL=DC$. Thus, $$BE^2=CD\cdot AE-AE\cdot AD=EA\cdot EL,$$thus $BE$ is tangent to $(ABL)$. As $BD$ is the angle bisector of $\triangle CDL$, we also have $BD$ the perpendicular bisector of $LC$. Hence, $$\measuredangle EBA=\measuredangle BLD=\measuredangle DCB,$$we are done.

We prove the converse, as such a point $E$ is clearly unique. Let $E'$ be the reflection of $E$ over $\overline{BD}$ and define $A'$ similarly. Then $$AE \cdot ED + BE^2 = DE' \cdot E'A' + BE'^2.$$Because $\overline{BE'}$ is tangent to $(A'BC)$ by the angle condition, $$DE' \cdot E'A' + BE'^2 = DE' \cdot E'A' + E'A' \cdot E'C = EA \cdot CD,$$as needed.

Reflect $C$ about $BD$ to get $C' \in \overrightarrow{DA}$. Then the condition becomes $EA \cdot EC' = EB^2$, and what is to be proven is the Alternate Segment Theorem, so we're done.