Prove that in RT△ABD, AD=2 AB=2√3 Pythagorean theorem: BD=4.

In RT△ABC, AB=2√3 BC=6 is obtained from Pythagorean theorem: AC=4√3.

AD/BC = DE/EB = AE/EC = 2/6 = 1/3

∴BE=3 EC=3√3

Is it? +EC？ = 9+27 = 36 = BC?

BE, EC and BC become Pythagoras numbers.

AC⊥BD……①

∵PA⊥ Bottom ABCD ∴ PA ⊥ BD...②

From ①② BD⊥ flat packaging

On the other hand, according to the judgment theorem of plane PBD⊥ plane PAC, BD is on the plane PBD.