AoPS - Problem

Problem

Source: 2022 Grand Duchy of Lithuania, MC p3 (Baltic Way TST) https://artofproblemsolving.com/community/c1321893_grand

Tags: geometry, Concyclic, isosceles

parmenides51

17.11.2022 23:36

The center $O$ of the circle $\omega$ passing through the vertex $C$ of the isosceles triangle $ABC$ ($AB = AC$) is the interior point of the triangle $ABC$. This circle intersects segments $BC$ and $AC$ at points $D \ne C$ and $E \ne C$, respectively, and the circumscribed circle $\Omega$ of the triangle $AEO$ at the point $F \ne E$. Prove that the center of the circumcircle of the triangle $BDF$ lies on the circle $\Omega$.

Let $AB \cap (AEF)= H$. Then $\angle BHF= \angle FEC=\angle FDC \Rightarrow BDFH$ is cyclic. Also $\angle EOD=2 \angle C = 180- \angle A = \angle EOH \Rightarrow O, D, H$ are collinear. Now let $ED \cap (BDFH) = I$. Since $\frac{\angle HEF}{2}=\frac{\angle HOF}{2}=\frac{\angle DOF}{2} = \angle DEF$, we have $IH=IF$. Also $\angle DIF + 2\angle IDF=\angle EAF + 2 \angle ECF = 180 - \angle EOF + 2 \angle ECF=180$, therefore we have $ID=IF$ as well, meaning $I$ is the required center lying on the circle $\omega$.