Angle EBA is equal to Angle DCB

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Angle EBA is equal to Angle DCB

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Source: Baltic Way 2011

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WakeUp

#1 h Nov 6, 2011, 1:25 PM • 3 Y

Y by anantmudgal09, Adventure10, Mango247

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 $\angle EBA=\angle DCB$ .

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newsun

#2 h Nov 6, 2011, 1:58 PM • 3 Y

Y by IstekOlympiadTeam, Adventure10, Mango247

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.

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256 posts

#3 h Oct 4, 2020, 6:50 AM

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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.

This post has been edited 1 time. Last edited by Kimchiks926, Oct 13, 2020, 1:15 PM
Reason: typo

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#4 h Aug 13, 2021, 9:14 PM

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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.

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#5 h Sep 13, 2022, 5:18 PM

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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.

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#6 h Dec 18, 2024, 5:51 PM

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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.

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#7 h Apr 10, 2025, 9:38 AM

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I had following simple solution. If we lock the points $B, E, D,C$ , then there is will be unique $A$ which requires the given condition. So let's assume that $\angle EBA=\angle DCB$ , and prove the given condition. Let $B'$ be the reflection of $B$ over $E$ . Let $F=(ABB') \cap AE$ . $G=BD \cap B'F$ .
Let $H$ be the midpoint of $BG$ . So it's clear that $2*EH=BD+DG$ . On the other hand we have: $\triangle FDG \sim \triangle FEH \sim \triangle CDB$ . Which means that $2*EH=BD+DG$ $\implies$ $2*EF=DF+DC$ $\implies$ $CD-DE=ED+DF=EF$ . On the other hand by Thales theorem: $BE*BE'=AE*EF$ $\implies$ $BE^2=AE(CD-DE)$ , and conclusion follows.

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