$解: (1)∵∠ACB=90°$
$∴AB=\sqrt{AC^2+BC^2}=10$
$∵D是AB的中点$
$∴DC=DB=DA=\frac 12AB=5$
$∴∠B=∠DCB$
$∵△ABC≌△ FDE$
$∴∠FDE=∠B∴∠FDE=∠DCB$
$∴DG//BC$
$∴∠AGD=∠ACB=90°$
$∴DG ⊥AC$
$∵DC=DA$
$∴G是AC的中点$
$∴CG=\frac 12AC=\frac 12×8=4$
$∴DG=\sqrt{CD^2-CG^2}=3$
$∴S_{△DCG}=\frac 12×CG×DG=\frac 12×4×3=6$
$ $
