The length of the hypotenuse of an isosceles right triangle is $\sqrt{2}$ times the length of each leg.  Therefore, $BD=\frac{BC}{\sqrt{2}}=\frac{2}{\sqrt{2}}\cdot\left(\frac{\sqrt{2}}{\sqrt{2}}\right)=\frac{2\sqrt{2}}{2}=\sqrt{2}$ units.  Applying the same rule to triangle $ABD$, we find that $AB=BD/\sqrt{2}=\sqrt{2}/\sqrt{2}=1$ unit.  The perimeter of quadrilateral $ABCD$ is $AB+BD+CD+DA=1+2+\sqrt{2}+1=4+\sqrt{2}$ units.
Each triangle in this figure is an isosceles right triangle. The length of $\overline{BC}$ is 2 units. What is the number of units in the perimeter of quadrilateral $ABCD$? Express your answer in simplest radical form.

[asy]
draw((0,0)--(10,0)--(0,10)--cycle);
draw((10,0)--(0,10)--(10,20)--cycle);
draw((1,0)--(1,1)--(0,1));
draw((1,11)--(2,10)--(1,9));
label("$A$",(0,0),S);
label("$B$",(10,0),S);
label("$C$",(10,20),N);
label("$D$",(0,10),W);

[/asy]