\because AB=AD=\sqrt{2} and point O is the midpoint of BD

\therefore \triangle ABD is a an isosceles triangle and

Using Pythagorean Theorem

AO = \sqrt{AB^2-(\dfrac{BD}{2})^2 } =\sqrt{2-1}=1

On the other hand,

\because CB=CD=BD=2

\therefore \triangle BCD is an equilateral triangle and

OC\perp BD

Using Pythagorean Theorem

OC = \sqrt{BC^2-(\dfrac{BD}{2} )^2} =\sqrt{4-1}=\sqrt{3}

For three sides of \triangle AOC ,

CA=2, OC=\sqrt{3}, AO=1,

CA^2=OC^2+AO^2

Therefore, \triangle AOC is a right triangle

Since AO is perpendicular to two intersecting lines in the plane of BCD,

AO \perp plane BCD