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.