Prove that chords are equal

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Prove that chords are equal

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Source: Argentina IMO TST 2006 problem 3

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#1 h Aug 27, 2009, 7:50 PM • 3 Y

Y by Adventure10, Mango247, Mango247

In a circumference with center $O$ we draw two equal chord $AB=CD$ and if $AB \cap CD =L$ then $AL>BL$ and $DL>CL$
We consider $M \in AL$ and $N \in DL$ such that $\widehat {ALC} =2 \widehat {MON}$
Prove that the chord determined by extending $MN$ has the same as length as both $AB$ and $CD$

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saif

#2 h Aug 27, 2009, 8:44 PM • 1 Y

Y by Adventure10

if possible can you draw it to see clearly the exercise
thank you very much.

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#3 h Feb 28, 2011, 5:01 PM • 2 Y

Y by Adventure10, Mango247

Let $OM$ and $ON$ cut line segment $BC$ at $E$ and $F$ respectively.And let $P$ and $Q$ be projections of $O$ to $AB$ and $CD$ respectively. $AB=CD$ and $ABCD$ -cyclic implies that $ABCD$ is isosceles trapezoid. $BL=CL \implies AL=DL \implies \angle BAD=\angle CDA = \angle DCB = \angle ABC$ . $OP \perp AB$ and $OQ \perp CD$ implies that $AP=PB$ and $QD=QC;AL=DL$ implies that $PL=QL \implies \angle LPQ = \angle LQP= \angle ABC$ which implies that $BC || PQ ;OL \perp PQ \implies \angle MOP= \angle NOL$ and $OL \perp BC \implies \angle DCB = \angle NOE \implies CENO$ -cyclic and similarly we get that $BFMO$ is cyclic and from $OB=OC$ we get that $\angle OBC= \angle OCB \implies \angle OCN =\angle OBM=\angle OEN=\angle OFM$ which implies that $MEFN$ is cyclic $\implies \angle OFE=\angle OMN$ and $\angle MOP=\angle NOL$ implies that $\angle OFE=\angle OMP \implies \angle OMP=\angle OMN \implies \angle OMX= \angle OMB$ if $MN$ cuts circle at $X$ and $Y$ . Let angle bisector of $\angle BOX$ cut $XM$ at $K$ and let $WLOG$ $K$ lies between
$X$ and $M \implies \angle OMX+\angle KOM =\angle OKX=\angle OKB$
$\implies \angle OKB >\angle OMX=\angle OMB$ which is contradiction thus we get that $M=K \implies MB=MX \implies AXBY$ is isosceles trapezoid which implies that $XY=AB$ .
_____________________________________________________
by Elshad Mustafayev[Sumgayit]

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#4 h Jan 24, 2013, 5:27 PM • 2 Y

Y by Adventure10, Mango247

It's enough to prove that $O$ is center of the excircle of $LMN$ and this is obvious because $2MON=ALC$ and $DLO=OLA$

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