Since two opposite of angles of quadrilateral $ABCD$ sum up to $180^{\circ}$, we know that it is cyclic. By Ptolemy's Theorem, we have $BD \cdot AC = AB \cdot CD + AD \cdot BC$. We know that $\Delta ADC$ is an isosceles right triangle as $AD = CD$ and $\angle ADC = 90^{\circ}$, so $AC = \sqrt{2} AD$. Substituting $AD$ for $CD$ and $AC$ we get $BD \cdot (\sqrt{2} AD) = AD \cdot (AB + BC)$ and cancelling out $AD$ we are left with $\sqrt{2} BD = AB + BC = 2021$. Thus BD = $\frac{2021}{\sqrt{2}} = \frac{2021\sqrt{2}}{2}$.