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
Orestis_Lignos
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.
nik_1oo1
14.06.2023 09:30
[asy][asy]
size(8cm);
//init
pair A = (-0.8, -0.6);
pair B = (0.8, -0.6);
pair C = (0.6, 0.8);
pair D = (0, -1);
pair E = (0.75, -0.75);
pair F = (0.6, -0.8);
pair O = (0, 0);
pair K = (0.6, -0.6);
//draw
draw(unitcircle);
draw(A--B--C--cycle);
draw(circumcircle(B, O, D));
draw(O--B);
draw(O--C);
draw(O--E);
draw(B--D);
draw(E--A, dashed);
draw(E--C, dashed);
draw(O--D); draw(D--E); draw(C--F);
markscalefactor=0.05;
draw(anglemark(E,O,B));
draw(anglemark(F, C, B));
draw(anglemark(E, D, B));
//label
dot("$A$", A, dir(180));
dot("$B$", B, dir(10));
dot("$C$", C, dir(50));
dot(“$D$”, D, dir(270));
dot(“$E$”, E, dir(0));
dot(“$F$”, F, dir(320));
dot(“$K$”, K, dir(30));
dot(“$O$”, O, dir(140));
[/asy][/asy]
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.