Suppose that the bisectors from $ B$ and $ D$ are parallel. The bisector of $ \angle ADC$ intersects $ AB$ in $ E$ and the bisector of $ \angle ABC$ intersects $ DC$ in $ F$. $ \angle DEB=A+\frac {D} {2}$ and $ \angle FBE=\frac {B} {2}$ $ DE \parallel FB \Rightarrow A+\frac {D} {2}+\frac {B} {2}=180^{\circ}\Rightarrow A=C=90^{\circ}$ Finally $ AB^2+AD^2=BC^2+CD^2=4R^2$

If the bisectors of B and D are parallel, doesnt that mean $ \angle B=\angle D=90$? Then, $ AB^2+BC^2=CD^2+AD^2=AC^2$.