DHA=∠ABD=45°,故②正确;③由②知:A、B、C、D、H五点共圆,则∠BAH=∠BDH;又∵∠ABD=∠DBG=45°,∴△ABM∽△DBG,得AM:DG=AB:BD=1:,即DG=AM;故③正确;④过H作HN⊥CD于N,连接NG;若BH平分∠DBG,且BH⊥DG,易知:BH垂直平分DG;得DE=EG,H是DG中点,HN为△DCG的中位线;设CG=1,则:HN=,EG=DE=,DC=BC=+1;易证得△BEC∽△HEN,则:BE:EH=BC:HN=2+2,即EH=;∴HE•BH=BH•=4-2,即BE•BH=4;∵∠DBH=∠CBE,且∠BHD=∠BCE=90°,∴△DBH∽△CBE,得:DB•BC=BE•BH=4,即BC2=4,得:BC2=4,即正方形ABCD的面积为4;故④正确;因此四个结论都正确,故选D11.D解:(1)连接HE,FC,延长HF交AD于点L,∵BD为正方形ABCD的对角线,∴∠ADB=∠CDF=45°.∵AD=CD,DF=DF,∴△ADF≌CDF.∴FC=AF,∠ECF=∠DAF.∵∠ALH+∠LAF=90°,∴∠LHC+∠DAF=90°.∵∠ECF=∠DAF,∴∠FHC=∠FCH,∴FH=FC.∴FH=AF.(2)∵FH⊥AE,FH=AF,∴∠HAE=45°.(3)连接AC交BD于点O,可知:BD=2OA,∵∠AFO+∠GFH=∠GHF+∠GFH,∴∠AFO=∠GHF.∵AF=HF,∠AOF=∠FGH=90°,∴△AOF≌△FGH.∴OA=GF.∵BD=2OA,∴BD=2FG.(4)延长AD至点M,使AD=DM,过点C作CI∥HL,则:LI=HC,根据△MEC≌△MIC,可得:CE=IM,同理,可得:AL=HE,∴HE+HC+EC=AL+LI+IM=AM=8.∴△CEM的周长为8,为定值.故(1)(2)(3)(4)结论都正确.故选D.12.A
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