equal angles, interior point of triangle, concyclic

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Source: Czech-Polish-Slovak Match 2017 day 1 P2

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#1 h Sep 28, 2017, 3:14 PM • 1 Y

Y by Adventure10

Let ${\omega}$ be the circumcircle of an acute-angled triangle ${ABC}$ . Point ${D}$ lies on the arc ${BC}$ of ${\omega}$ not containing point ${A}$ . Point ${E}$ lies in the interior of the triangle ${ABC}$ , does not lie on the line ${AD}$ , and satisfies ${\angle DBE =\angle ACB}$ and ${\angle DCE = \angle ABC}$ . Let ${F}$ be a point on the line ${AD}$ such that lines ${EF}$ and ${BC}$ are parallel, and let ${G}$ be a point on ${\omega}$ different from ${A}$ such that ${AF = FG}$ . Prove that points ${D,E, F,G}$ lie on one circle.

(Slovakia)

This post has been edited 1 time. Last edited by parmenides51, Sep 28, 2017, 3:16 PM
Reason: country proposer added

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soryn

#2 h Jan 28, 2018, 7:58 AM • 2 Y

Y by Adventure10, Mango247

Any solution please?

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#3 h Jan 28, 2018, 8:58 AM • 1 Y

Y by Adventure10

WLOG $AB \geq AB$ . By an easy angle chase $\angle BEC=2 \angle A$ so $BOEC$ is cyclic where $O$ is the circumcentre of $\triangle ABC$ .

First we prove $FOED$ is cyclic:
$\angle ODF=\angle ODA=\frac{\pi-\angle DOA}{2}=\frac{\pi}{2}-\angle DCA=\frac{\pi}{2}-(\angle DCB+\angle C)=\frac{\pi}{2}-\angle C-\angle A+\angle DAC$
$\angle OEF=\angle OEB-\angle FEB=\angle OCB-\angle EBC=\frac{\pi}{2}-\angle A-\angle EBD+\angle CBD=\frac{\pi}{2}-\angle A-\angle C+\angle DAC$
So $\angle OEF=\angle ODF \implies FOED$ is cyclic.

Next we prove $OFDG$ is cyclic:
$\angle OFG=\angle OFA=\frac{\pi}{2}-\angle DAG=\frac{\pi}{2}-\frac{\angle DOG}{2}=\angle ODG$

Hence we have $OFDGE$ cyclic as desired.

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soryn

#4 h Jan 28, 2018, 10:14 AM • 2 Y

Y by Adventure10, Mango247

Thank you!!!

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