$解：(1)△ADG∽△ACD，△CDG∽△CAD$
$∵四边形ABCD为矩形$
$∴∠ADC=90°$
$∵DG⊥AC$
$∴∠AGD=∠DGC=∠ADC=90°$
$又∠DAG=∠DAC，∠DCG=∠DCA$
$∴△ADG∽△ACD，△CDG∽△CAD$
$(2)∵△ADG∽△ACD，△CDG∽△CAD$
$∴△ADG∽△DCG$
$∴\frac {AG}{DG}=\frac {DG}{CG}$
$∵AG=6，CG=12$
$∴DG=6\sqrt 2$
$∴S_{矩形ABCD}=2S_{△ADC}=2×\frac 12×AC · DG=2×\frac 12×(6+12)×6\sqrt 2=108\sqrt 2$
