∴AB∥HF∥DC∥GN，

Let AC and FH intersect at p, CD and HG intersect at q,

∴△PFC, △QCG and △NGE are regular triangles.

F and g are the midpoint of BC and CE respectively.

∴bf=mf= 12ac= 12bc,cp=pf= 12ab= 12bc

∴CP=MF,CQ=BC,QG=GC=CQ=AB，

∴S 1= 12S,S3=2S，

∫s 1+S3 = 10

∴ 12S+2S= 10，

∴S=4.

So the answer is: 4.