Construct the circumcircle to triangle $ABC$ in which $AB = 5 \; cm$, $BC = 8 \; cm$ and $\angle ABC = 60^\circ$.
Construct cyclic quadrilateral $ABCD$ such that $D$ is equidistant from $B$ and $C$.
Using the given data, construct $\triangle ABC$ with $AB = 5 \; cm$, $BC = 8 \; cm$ and $\angle ABC = 60^\circ$.
Draw the perpendicular bisectors of sides $AB$ and $AC$ which intersect each other at point $O$.
With $O$ as center and $OA$ as radius draw a circle which passes through the vertices of $\triangle ABC$.
This circle is the required circumcircle to $\triangle ABC$.
Since point $D$ is equidistant from the points $B$ and $C$, $D$ lies on the perpendicular bisector of $BC$.
The point where the perpendicular bisector of $BC$ intersects the circumcircle is the point $D$.
$ABCD$ is the required cyclic quadrilateral.