Let $ABC$ be a triangle with circumcircle $\Gamma$ and circumcentre $O$. The circle with centre on the line $AB$ and passing through the points $A$ and $O$ intersects $\Gamma$ again in $D$. Similarly, the circle with centre on the line $AC$ and passing through the points $A$ and $O$ intersects $\Gamma$ again in $E$. Prove that $BD$ is parallel with $CE$.
Problem
Source: Baltic Way 2022/11
Tags: geometry
VicKmath7
13.11.2022 00:18
Easy angle chasing gives that $\angle ADO=\gamma$. Let $DO \cap \Gamma=R$; then $\angle DAR=90^o$, so $\angle DBA= \angle DRA =90^o-\gamma$. Thus $\angle DBC=90^o-\gamma+\beta=\angle BCE$ (the last equality is obtained similarly), so $DBEC$ is an isosceles trapezoid.
gghx
24.11.2022 12:21
Note that $AO=EO\implies \angle OEY=\angle OAY=\angle OCY\implies OYEC$ is cyclic. Thus, $OY=YE\implies \angle ECA=\angle YEO=\angle OAC\implies AO\parallel EC$. Similarly, $AO\parallel BD$, so we are done.
Radin.A
16.04.2024 17:58
Denote the circumcentre of the first and second circle as F , K Just note that OKEC , OFDB are cyclic since $\angle CKE= \angle COE$ And $\angle BOD = \angle BFD$ so BD, CE are parallel to AO.