Solution:

The quadrilateral EFGH is similar to the original isosceles trapezoid ABCD for the following reasons:

E is the midpoint of OD and f is the midpoint of OC.

∴EF is the center line of △COD.

∴EF= 1/2CD,EF‖CD，

∴∠OEF=∠ODC,∠OFE=∠OCD

Similarly:

HF is the center line of △BOC.

∴HF= 1/2BC,HF‖BC

∴∠OFH=∠OCB,∠OHF=∠OBC

GH is the center line of delta △AOB.

∴GH= 1/2AB,GH‖AB

∴∠OHG=∠OBA,∠OGH=∠OAB

GE is the center line of △AOD.

∴GE= 1/2AD,GE‖AD

∴∠OGE=∠OAD,∠OEG=∠ODA

∴∠GEF=∠OEF+∠OEG=∠ODC+∠ODA=∠ADC

∠EFH =∠OFE+∠OFH =∠ OCB+∠ DCB.

∠FHG =∠OHF+∠OHG =∠OBC+∠ Oba =∠CBA.

∠HGE =∠OGH+∠OGE =∠OAB+∠OAD =∠ bud.

The corresponding edges are proportional and the corresponding angles are equal.

∴ The quadrilateral EFGH is similar to the original isosceles trapezoid ABCD.