∵四边形AEFG∽四边形ABCD
∴AE:AB=AG:AD
又∵AB=a,AD=BC=2a,AE=2/3a
∴2/3a:a=AG:2a
∴AG=4/3a
(2)证明：∵∠GAD=∠GAE-∠DAE,∠EAB=∠BAD-∠DAE
又 ∵在矩形AEFG和矩形ABCD中,∠GAE=∠BAD=90°
∴∠GAD=∠EAB
又∵AE:AB=AG:AD ,即AE:AG=AB:AD
∴△ABE∽△ADG
∵矩形AEFG∽矩形ABCD
∴ S矩形ABCD：S矩形AEFG=(AB:AE)^2=（a:2/3a)^2=9/4
又∵S矩形ABCD=630cm平方
∴S矩形AEFG=S矩形ABCD/(9/4)=630×4/9=280cm平方