AoPS - Problem

Source: 2021 Oral Moscow Geometry Olympiad grades 8-9 p4

Tags: geometry, cyclic quadrilateral, circumcircle

Ya_pank

17.05.2021 15:51

On the diagonal $AC$ of cyclic quadrilateral $ABCD$ a point $E$ is chosen such that $\angle ABE = \angle CBD$. Points $O,O_1,O_2$ are the circumcircles of triangles $ABC, ABE$ and $CBE$ respectively. Prove that lines $DO,AO_{1}$ and $CO_{2}$ are concurrent.

Let $F$ be the antipode of point $D$ in $(ABC)$. I claim that lines $DO$, $AO_1$, $CO_2$ pass through point $F$. $$\angle ABE=\angle DBC=\angle CAD$$which means, that $DA$ is tangent to $(ABE)$. We have that $\angle FAD=90^{\circ}$, since $DF$ is diameter and $\angle O_1AD=90^{\circ}$, since $DA$ is tangent to $(ABE)$. This means that points $F$, $O_1$, $A$ are collinear. Analogously (but slightly more tricky) we have that points $F$, $O_2$, $C$ are collinear. This means that lines $DO$, $AO_1$, $CO_2$ are concurrent at point $F$, as desired.