Reference card method: let the midpoint of BD be m and connect AM and CM.

It is easy to know cm⊥bd am⊥bd (the midline of the bottom of an isosceles triangle is also high).

Then ∠AMC is the dihedral angle between plane ABD and plane BCD.

am = √ 2a/2。

CM=√ 2a/2

AC=a

That is, △AMC satisfies Pythagorean theorem

So ∠ AMC = 90 degrees.

So plane ABD is perpendicular to plane BCD.