$解:(2)連接OD$
$則S_{△BDF}=S_{△OBD}$
$=S_{△BOA}-S_{△OAD}$
$=\frac{1}{2}×8-\frac{1}{2}×2=3$
$(3)設(shè)點(diǎn)D(m,\frac{2}{m}),則點(diǎn)B(4m,\frac{2}{m})$
$∵點(diǎn)G與點(diǎn)O關(guān)于點(diǎn)C對(duì)稱(chēng)點(diǎn),∴G(8m,0)$
$易得點(diǎn)E(4m,\frac{1}{2m})$
$通過(guò)點(diǎn)D、E的坐標(biāo)可求得直線DE的函數(shù)表達(dá)式為$
$y=-\frac {1}{2m^{2}}x+\frac {5}{2m}$
$令y=0,x=5m,則F(5m,0),∴FG=8m-5m=3m$
$又BD=4m-m=3m=FG,且FG//BD$
$∴四邊形BDFG為平行四邊形 $
