AoPS - Problem

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

Tags: Concyclic, equal angles, geometry, circle

parmenides51

28.09.2017 18:14

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)

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.