（1）證明：如圖1，作∠BAP=∠DAE，AP交BD於P，設∠CBD=α，∠CAD=β，∵∠ADB=∠CAD+∠ABD，∠APE=∠BAP+∠ABD，∴∠APE=∠ADE，AP=AD．∵AC⊥BD∴∠PAE=∠DAE=β，∴∠PAD=2β，∠BAD=3β．∵∠BAD=3∠CBD，∴3β=3α，β=α．∵AC⊥BD，∴∠ACB=90°-∠CBE=90°-α=90°-β．∵∠ABC=180°-∠BAC-∠ACB=90°-β，∴∠ACB=∠ABC，∴△ABC為等腰三角形；（2）2MH=FM+CD．證明：如圖2，由（1）知AP=AD，AB=AC，∠BAP=∠CAD=β，∴△ABP≌△ACD，∴∠ABE=∠ACD．∵AC⊥BD，∴∠GDN=90°-β，∵GN=GD，∴∠GND=∠GDN=90°-β，∴∠NGD=180°-∠GND-∠GDN=2β．∴∠AGF=∠NGD=2β．∴∠AFG=∠BAD-∠AGF=3β-2β=β．∵FN平分∠BFM，∴∠NFM=∠AFG=β，∴FM∥AE，∴∠FMN=90°．∵H為BF的中點，∴BF=2MH．在FB上擷取FR=FM，連線RM，∴∠FRM=∠FMR=90°-β．∵∠ABC=90°-β，∴∠FRM=∠ABC，∴RM∥BC，∴∠CBD=∠RMB．∵∠CAD=∠CBD=β，∴∠RMB=∠CAD．∵∠RBM=∠ACD，∴△RMB∽△DAC，∴，∴BR=CD．∵BR=FB-FM，∴FB-FM=BR=CD，FB=FM+CD．∴2MH=FM+CD．