$解：?(1)△ADG∽△ACD，??△CDG∽△CAD?$
$∵四邊形?ABCD?為矩形$
$∴?∠ADC=90°?$
$∵?DG⊥AC?$
$∴?∠AGD=∠DGC=∠ADC=90°?$
$又?∠DAG=∠DAC，??∠DCG=∠DCA?$
$∴?△ADG∽△ACD，??△CDG∽△CAD?$
$?(2)?∵?△ADG∽△ACD，??△CDG∽△CAD?$
$∴?△ADG∽△DCG?$
$∴?\frac {AG}{DG}=\frac {DG}{CG}?$
$∵?AG=6，??CG=12?$
$∴?DG=6\sqrt 2?$
$∴?S_{矩形ABCD}=2S_{△ADC}=2×\frac 12×AC · DG=2×\frac 12×(6+12)×6\sqrt 2=108\sqrt 2?$
