初中数学几何题求解~详情见下图~

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(1)证明:延长CM,交AB的延长线于E.
∠AME=∠AMC=90度;AM=AM;∠EAM=∠CAM.则⊿EAM≌ΔCAM,得EM=CM;AE=AC.
过点E作DM的平行线,交CB的延长线于N,则⊿CDM∽⊿CNE,EN/DM=EC/CM=2,即EN=2DM.
且∠N=∠ADB;又AB=AD,则∠ADB=∠ABD=∠EBN.故∠N=∠EBN,得EN=EB.
故:AC-AB=AE-AB=EB=EN=2DM.
(2)AB+AC=2DM.
证明:延长CM,交BA的延长线于E,同理可证:⊿EAM≌ΔCAM,EM=CM;AE=AC.
过点E作DM的平行线,交BD的延长线于N,则⊿CDM∽⊿CNE,EN/DM=EC/CM=2,即EN=2DM.
AB=AD,则∠B=∠ADB;故:∠N=∠ADB=∠B,得EN=EB.
所以,AB+AC=AB+AE=EB=EN=2DM.
(3)AB+AC=2DM.
证明:延长BA,交CM的延长线于E,则∠CAM=∠DAP;∠EAM=∠DAB.
又∠DAB=∠DAP,故∠CAM=∠EAM.
又AM=AM,∠AMC=∠AME=90度.则⊿CAM≌ΔEAM,得AE=AC;AM=CM,EC=2CM.
过点E作MD的平行线,交CD的延长线于N,则∠N=∠ADB;
AB=AD,则∠ADB=∠ABD.故∠N=∠ABD,得:EN=EB.
⊿ENC∽⊿MDC,则EN/DM=EC/CM=2,EN=2DM.
所以,AB+AC=AB+AE=EB=EN=2DM.