Given a parallelogram $ABCD$ with angle $A$ equal to $60^o$. Point $O$ is the the center of a circle circumscribed around triangle $ABD$. Line $AO$ intersects the bisector of the exterior angle $C$ at point $K$. Find the ratio $AO/OK$.
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Tags: ratio, geometry, parallelogram
Given a parallelogram $ABCD$ with angle $A$ equal to $60^o$. Point $O$ is the the center of a circle circumscribed around triangle $ABD$. Line $AO$ intersects the bisector of the exterior angle $C$ at point $K$. Find the ratio $AO/OK$.