We are given an acute $\triangle ABC$ with circumcenter $O$ such that $BC<AB$. The bisector of $\angle ACB$ meets the circumcircle of $\triangle ABC$ at a second point $D$. The perpendicular bisector of $AC$ meets the circumcircle of $\triangle BOD$ for the second time at $E$. The line $DE$ meets the circumcircle of $\triangle ABC$ for the second time at $F$. Prove that the lines $CF$, $OE$ and $AB$ are concurrent. Proposed by Petar Filipovski
Problem
Source: 2023 Junior Macedonian Mathematical Olympiad P4
Tags: geometry
10.06.2023 13:22
Let $EO$ intersect $AB$ at point $K$. We need to prove that points $F,K,C$ are collinear. Since $\angle ACK=\angle KAC=\angle A,$ it is enough to show that $\angle FCA=\angle A$. However, this dies to an easy angle-chase: $\angle FCA=180^\circ-\angle FDA=180^\circ-(\angle EDB+\angle BDA)=180^\circ-(\angle EOB+(180^\circ-\angle C))=\angle C-\angle EOB=\angle C-(\angle C-\angle A)=\angle A,$ as desired.
14.06.2023 09:30
We use standard notations for the angles $\alpha$, $\beta$ and $\gamma$. Let $EO\cap CF=K$. We will prove that $K\in AB$. $\angle AOC=2\beta$ $O\in S_{AC}$ and $E\in S_{AC}$ $\Rightarrow$ $\angle COE=\angle AOE=180^{\circ} -\beta$ $DEBO$ is cyclic $\Rightarrow \angle EDB=\angle EOB$ $DFBC$ is cyclic $\Rightarrow \angle FDB=\angle FCB$ $\Rightarrow \angle KOB = \angle KCB \Rightarrow KOCB$ is cyclic $\Rightarrow \angle KOC + \angle KBC = 180^{\circ}$ $180^{\circ} - \beta + \angle KBC = 180^{\circ}$ $\angle KBC = \beta$ $\angle CBA = \angle CBK = \beta \Rightarrow K\in AB$ and the problem is proved.