Let $D$ be the intersection of $CO$ with the circle. By definition $CD$ is a diameter as $O$ is the centre of the circle. $\angle CAD=90^{\circ}$ so $AD^{2}+AC^{2}=CD^{2}\Rightarrow AD^{2}+AC^{2}=constant$ But $AD=CB$ because $CD$ and $AB$ are parallel so we have $AD^{2}+CB^{2}=constant$. Note: $AD=CB$ because if $\angle DCB=\alpha$ we have $\angle CBA=180^{\circ}-\alpha \Rightarrow \angle ADC=\alpha$ using $CD\parallel AB$ and $ABCD$ is a cyclic quad in that order. So $ABCD$ is an isosceles trapezium and $AD=CB$.