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.