Školjka

%V0 Let $ABC$ be an acute-angled triangle such that $\angle ABC<\angle ACB$, let $O$ be the circumcenter of triangle $ABC$, and let $D=AO\cap BC$. Denote by $E$ and $F$ the circumcenters of triangles $ABD$ and $ACD$, respectively. Let $G$ be a point on the extension of the segment $AB$ beyound $A$ such that $AG=AC$, and let $H$ be a point on the extension of the segment $AC$ beyound $A$ such that $AH=AB$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if $\angle ACB-\angle ABC=60^\circ$. comment Official version Let $O$ be the circumcenter of an acute-angled triangle $ABC$ with ${\angle B<\angle C}$. The line $AO$ meets the side $BC$ at $D$. The circumcenters of the triangles $ABD$ and $ACD$ are $E$ and $F$, respectively. Extend the sides $BA$ and $CA$ beyond $A$, and choose on the respective extensions points $G$ and $H$ such that ${AG=AC}$ and ${AH=AB}$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if ${\angle ACB-\angle ABC=60^{\circ }}$. Edited by orl.