A point $D$ is taken on the arc $BC$ of the circumcircle of triangle $ABC$ which does not contain $A$. A point $E$ is taken at the intersection of the interior region of the triangles $ABC$ and $ADC$ such that $m(\widehat{ABE})=m(\widehat{BCE})$. Let the circumcircle of the triangle $ADE$ meets the line $AB$ for the second time at $K$. Let $L$ be the intersection of the lines $EK$ and $BC$, $M$ be the intersection of the lines $EC$ and $AD$, $N$ be the intersection of the lines $BM$ and $DL$. Prove that $$m(\widehat{NEL})=m(\widehat{NDE})$$
Problem
Source: 2021 Turkey TST P3
Tags: geometry, circumcircle, Turkey, Pascal s therem
23.05.2021 22:21
Nice solution by Burak Buğra Önder Let $\angle KED=\angle KAD=\beta$. We have $\angle BCD=\angle BAD=\angle KAD=\beta$. Then $L,E,C,D$ is cyclic. Let $\angle LDE=\angle LCE=\alpha$. Let $AD\cap BE=X$. We have $\angle EXD=\angle AXB=180^\circ-\angle BAX-\angle ABX=180^\circ-\beta-\angle BCE=180^\circ-\beta-\alpha$. Also, $\angle ECD=\alpha+\beta$. Then $E,C,D,X$ is cyclic. Then $X,L,D,C,E$ is cyclic. Apply Pascal theorem to $EECLDX$. $EC\cap DX=M$ and $CL\cap XE=B$. Then $EE\cap LD,M,B$ are collinear. Then $EE$ pass through $LD\cap MB=N$. Hence, $EN$ is a tangent. Thus, $\angle NEL=LDE$. This completes the proof.
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23.05.2021 23:45
Basically same as @above Note that $\angle LED = \angle KED = \angle KAD = \angle BAD = \angle BCD = \angle LCD \implies CDLE$ is cyclic. Let $F = AD\cap BE$. Now note that $\angle ABF = \angle ABE = \angle ECL = \angle EDL$ and $\angle BAF = \angle LED$ so this gives $\triangle ABF \sim \triangle EDL$. And so $\angle ELD = \angle AFB = \angle EFD \implies EFLD$ is cyclic. And so $CEFLD$ is cyclic. Now apply Pascal to $EEFDLC$ to get $EE\cap DL , B , M$ are collinear and so $EE\cap DL = N$, implying that $\angle NEL = \angle LDE = \angle NDE$, as desired. $\square$
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04.01.2022 09:04
$$T = AD\cap BE$$$\angle LED=\angle KAD=\angle BCD=\angle LCD$ $L,E,C,D$ Concyclic. $\angle BTD=\angle BAD+\angle ABE=\angle BCD+\angle BCE=\angle DCE$ $D,L,E,C,T$ concyclic.$EETDLC$ Pacsal theorem. $EN$ Tangent $(DLE)$. $ \angle NEL= \angle TEL + \angle NET = \angle TDL + \angle TDE = \angle LDE = \angle NDE $
04.01.2022 09:08
very nice problem.