Given the circle, midpoint $O(0,0)$ and radius $r$. The points $A(r,0),B(r\cos \beta,r\sin \beta),C(r\cos \gamma,r\sin \gamma),D(r\cos \delta,r\sin \delta)$. Here, we know $\beta+\delta=\gamma+180^{\circ} \quad (1)$. Slope of the line $AC\ :\ m_{AC}=\frac{\sin \gamma}{\cos \gamma -1}$, slope of the line $BD\ :\ m_{BD}=\frac{\sin \beta-\sin \delta}{\cos \beta-\cos \delta}$. $AC \bot BD \Rightarrow \sin \gamma(\sin \beta-\sin \delta)+(\cos \gamma -1)(\cos \beta-\cos \delta)=0$, $\sin \beta\sin \gamma -\sin \gamma\sin \delta + \cos \beta\cos \gamma -\cos \gamma \cos \delta -\cos \beta+\cos \delta=0$, $\cos \beta\cos \gamma +\sin \beta\sin \gamma =\sin \gamma\sin \delta +\cos \gamma \cos \delta +\cos \beta-\cos \delta \quad (2)$. Equation of the line $BC\ :\ x \sin \delta +y(1-\cos \delta)-r\sin \delta=0$. Distance of the point $O$ to this line $\frac{-r\sin \delta}{\sqrt{2-2\cos \delta}}$, half the length of $BC$ is $\frac{1}{2} \cdot r \cdot \sqrt{2-2\cos \beta\cos \gamma-2\sin \beta\sin \gamma}$. To prove: $\frac{-r\sin \delta}{\sqrt{2-2\cos \delta}}=\frac{1}{2} \cdot r \cdot \sqrt{2-2\cos \beta\cos \gamma-2\sin \beta\sin \gamma}$ or $\frac{-\sin \delta}{\sqrt{1-\cos \delta}}=\sqrt{1-\cos \beta\cos \gamma-\sin \beta\sin \gamma}$ or $\frac{\sin^{2}\delta}{1-\cos \delta}=1-\cos \beta\cos \gamma-\sin \beta\sin \gamma$ or $1+\cos \delta=1-\cos \beta\cos \gamma-\sin \beta\sin \gamma$. Using (2): $1+\cos \delta=1-(\sin \gamma\sin \delta +\cos \gamma \cos \delta +\cos \beta-\cos \delta)$. $\cos \delta=-\cos(\delta-\gamma) -\cos \beta+\cos \delta$. $0=-\cos(\delta-\gamma) -\cos \beta$. $\cos(\delta-\gamma)=-\cos \beta$: OK, see (1).