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Without loss of generality, let $\angle BAC=60^{\circ }$ and $|AB|>|CA|$. Let $D$ be the intersection of the angle bisector of $\angle BAC$ and the circumcircle of the triangle $ABC$. Point $D$ lies on the perpendicular bisector of $BC$. Let $P$ be the midpoint of $BC$. Since $|OC|=|OD|$ and $\angle COD=2\angle CAD=60^{\circ }$, triangle $COD$ is equilateral. $P$ is the midpoint of $OD$, since $OD\perp BC$. Since $|AH|=2|OP|$ we conclude that $|AH|=|OD|$. Also, we notice that the lines $AH$ and $OD$ are parallel, because they are both perpendicular to $BC$. Hence, $AHDO$ is a parallelogram. Since $|AO|=|DO|$, $AHDO$ is a rhombus, therefore, its diagonals $AD$ and $OH$ are perpendicular to each other.