(1)证明:∵在梯形ABCD中,AD∥BC,AB=DC,
∴∠B=∠C,
∵GF=GC,
∴∠GFC=∠C,
∴∠B=∠GFC,
∴AB∥GF,
又∵AE=GF,
∴四边形AEFG是平行四边形;(2)解:若四边形AEFG是矩形,则∠EFB=
∠FGC.1 2
证明如下:过G作GH⊥FC,垂足为H,
∵GF=GC,
∴∠FGH=
∠FGC,且∠FGH+∠GFC=90°,1 2
∵∠EFG=90°,
∴∠EFB+∠GFH=90°,
∴∠EFB=∠FGH,
∴∠EFB=
∠FGC.1 2