$证明:(1)在四边形ABCD中$
$∵E、F、G、H分别是AD、BC、BD、AC的中点,$
$∴FG= \frac {1}{2}CD,HE =\frac {1}{2}CD,FH=\frac {1}{2}AB,GE=\frac {1}{2}AB$
$∵AB = CD$
$∴FG=FH= HE= EG.$
$∴四边形EGFH是菱形$
$(2)解:在四边形ABCD中,G,F,H分别是BD、BC、AC的中点$
$∴GF //DC,HF//AB$
$∴∠GFB=∠DCB,∠HFC=∠ABC$
$∴∠HFC+∠GFB =∠ABC+∠DCB =90°$
$∴∠GFH=90°$
$∴菱形EGFH是正方形$
$∵AB=1$
$∴EG=\frac {1}{2}AB=\frac {1}{2}$
$∴正方形EGFH的面积=(\frac {1}{2})²=\frac {1}{4}$