Extend AB so that CD is perpendicular to BD, then CD is the required distance.

because

because

In a right-angled triangular ADC:

AC square =AD square +CD square

4CD squared =(AB+BD) squared +CD squared

4CD squared =(2+CD) squared +CD squared.

Square of CD -2CD-2=0

CD = 1+√3？

CD= 1-√3 (s)