geometry in a isosceles

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Source: 2nd National Women´s Contest of Mexican Mathematics Olympiad 2023 , problem 1 teams

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AlanLG

#1 h Jul 23, 2023, 4:38 AM

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Let $\triangle ABC$ such that $AB=AC$ , $D$ and $E$ points on $AB$ and $BC$ , respectively, with $DE\parallel AC$ . Let $F$ on line $DE$ such that $CADF$ it´s a parallelogram. If $O$ is the circumcenter of $\triangle BDE$ , prove that $O,F,A$ and $D$ lie on a circle.

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#2 h Jul 23, 2023, 5:05 AM

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Consider that $AD=EF,\angle ODA=\angle OEF,OD=OE$
Hence $\triangle ODA\cong \triangle OEF$
Which means that $\triangle ODE\sim\triangle OAF$
So $\angle ODF=\angle OAF\Rightarrow O,F,A,D$ are cyclic

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jayme

#3 h Jul 23, 2023, 2:50 PM

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Dear Mathlinkers,

another point of view by considerind the pivot theorem...

Sncerely
Jean-Louis

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#4 h Sep 2, 2023, 1:35 PM

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Since $OB=OD$ , $OF=AC=AB$ and $\angle ABO=90-\angle DEB=90-\angle DBE=\angle ODF$ we have $\triangle OBA\cong \triangle ODF$ , which gives that $\angle DFO=\angle BAO=\angle DAO$ , which completes the solution.

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gnoka

#5 h Nov 1, 2023, 7:07 AM

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Since $AB=AC, DE//AC$ we have $BD=DE$ ,then $OD$ is angle bisector of $\angle BDE$
from parallelogram ADFC we have $DF=AC=AB=AD+DB$
which means $DF=AD+2OD\cos\angle ODB$
then $DF\sin\angle ODB=AD\sin\angle ODB+OD\sin2\angle ODB$
So $DF\sin\angle ADO=AD\sin\angle ODE+OD\sin2\angle ADE$
According to Ptolemy's theorem, $O,F,A$ and $D$ lie on a circle.

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