In RT△ABE, ∠B=90, so ∠E=30.

AE=2AB=8，BE=√3AB=4√3

In RT△CDE, ∠E=30.

So CE=2CD=4, DE=√3CD=2√3.

S△ABE= 1/2×AB×BE=8√3

S△CDE= 1/2×CD×DE=2√3

S quadrilateral ABCD=S△ABE-S△CDE=6√3

(8)①EF is the median vertical line of AD, so AE=DE.

Therefore ∠EAD=∠EDA

②F is on the vertical line of AD, so AF=DF.

∠DAF=∠ADF

AD divides ∠BAC equally, so ∠DAF=∠DAC.

So ∠ADF=∠DAC, DAC.

③ Starting from ①, ∠EAD=∠EDA.

∠EAC=∠EAD-∠DAC

∠EDA is the external angle of △ABD, so ∠B=∠EDA-∠DAF.

Because ∠DAC=∠DAF, ∠ EAC = ∠ B.

(7)①AE=CF, so AE+EF=CF+EF.

That is AF=CE.

Bf⊥ac de⊥ac, so ∠BFA=∠DEC=90.

There is AB=CD again

So △ ABF △ CDE (HL), BF=DE.

In △ reorganization and △DGE,

∠BGF=∠DGE，∠BFG=∠DEG=90，BF=DE

So △ bgf △ dge, BG=DG.

②AE=CF, so AE-EF=CF-EF.

That is AF=CE.

∠AFB=∠CED=90，AB=CD

So △ ABF △ CDE, BF=DE.

At △BFG and △DEG

∠BFG=∠DEG=90，∠BGF=∠DGE，BF=DE

So △ bfg △ deg, BG=DG.

The conclusion still holds.