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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 $\angle ACB-\angle ABC=60^\circ$.

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This is an interesting solution. slither io