$证明:∵四边形ABCD是正方形，\ $ $∴AB=AD=BC=CD,∠BAD=90°.\ $ $∵FH⊥CD,DE⊥BE,\ $ $∴∠FHD=90°=∠DEB=∠BAD.\ $ $又∠AGB=∠DGE,$ $∴∠ABG=∠EDG.\ $ $∵∠ADE+∠FDH=90°,$ $∠FDH +∠DFH=90°,\ $ $∴∠EDA=∠DFH,\ $ $∴∠ABG=∠DFH,\ $ $又DF=BG，\ $ $∴△ABG≌△HFD(AAS),\ $ $∴FH=AB.$
$证明:∵AB=BC=HF,$ $∠FHM=∠C=90°,$ $∠BMC=∠FMH,\ $ $∴△BCM≌△FHM(AAS),$ $∴BM=MF,$ $∵点O是BD的中点,$ $∴BO=DO,\ $ $∴OM=\frac{1}{2}DF,$ $∴OM=\frac{1}{2}BG.$
$解:如图，过点F作FN⊥直线BC于N.\ $ $∵AB=6,AG=2,\ $ $∴BG=AB²+AG²= \sqrt{36+4}=2 \sqrt{10},\ $ $∴DF=BG=2 \sqrt{10}.$ $∵AB=AD=6,∠BAD=90°,\ $ $∴BD=\sqrt{2}AB=6\sqrt{2}.\ $ $∵△ABG≌△HFD,\ $ $∴DH=AG=2,FH=AB=6,$ $∴CH=4.\ $ $∵FH⊥CD,FN⊥BC,∠DCN=90°,\ $ $∴四边形FHCN是矩形，\ $ $∴CH=FN=4,FH=CN=6,$ $∴BN=12,\ $ $∴BF=\sqrt{BN²+FN²}=\sqrt {144+16} =4 \sqrt{10}.\ $ $∵BE⊥EF,点O是BD的中点,BM=MF,\ $ $∴EO=\frac{1}{2}BD=3\sqrt{2},$ $EM=\frac{1}{2}BF=2 \sqrt{10}.\ $ $∵OM=\frac{1}{2}DF= \sqrt{10},\ $ $∴△OEM的周长= \sqrt{10}+3 \sqrt{2}+2 \sqrt{10}=3\sqrt{2}+3 \sqrt{10}.\ $
$解:如图(1)，连接GH.\ $ $由(1)得AG=BH,AG//BH,∠B=90°,$ $∴四边形ABHG是矩形,\ $ $∴GH=AB=6.\ $ $AC= \sqrt{AB²+BC^{3} }=10.\ $ $①如图(1),当四边形EGFH是矩形时，\ $ $∴EF=GH=6.\ $ $∵AE=CF=t,\ $ $∴EF=10-2t=6,\ $ $∴t=2.\ $ $②如图(2),当四边形EGFH是矩形时，\ $ $∵EF=GH=6,AE=CF=t,\ $ $∴EF=t+t-10=2t-10=6,$ $∴t=8.\ $ $综上,四边形EGFH为矩形时t=2或t=8.$
$解:如图(3),M和N分别是AD和BC的中点,连接AH、CG、GH,AC与GH交于O.\ $ $∵四边形EGFH为菱形,$ $∴GH⊥EF,OG=OH,OE=OF,$ $∴OA=OC,AG=AH,$ $∴四边形AGCH为菱形,$ $∴AG=CG.$ $设AG=CG=x,$ $则DG=8-x,$ $由勾股定理,$ $得CD²+DG²=CG²,$ $即6²+(8-x)²=x²,$ $解得x=\frac {25}{4},$ $∴MG=\frac {25}{4}-4=\frac {9}{4},$ $即t=\frac {9}{4},$ $∴当t=\frac {9}{4}时,四边形EGFH为菱形.$
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