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.