Aptitude : Which digit does E represent?

By given condition, we have 
A+B+C = 13...(1) 
C+D+E = 13...(2) 
E+F+G = 13...(3) 
G+H+I = 13...(4) 
Add up (1)+(2)+(3)+(4), we get  
52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I) 
= C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45 
 
Hence,C+E+G = 52 - 45 = 7...(5) 
 
By (5), we have E = 7-C-G <= 7-3 = 4 (since, the smallest C,G are 1 or 2)  
By (2)-(5)., we get D-G = 6, or D = G+6...(6)  
Since D,G are between 1 and 9, 
so D= 7, 8,or 9, and we have G=1,2 or 3 
 
By (3)-(5)., we get F-C = 6, or F = C+6...(7)  
Since F,C are between 1 and 9, 
so F= 7, 8,or 9, and we have C=1,2 or 3 
 
By (6) ,(7) & (5): 
Case(i) When G =1, D = G+6 =7,  
and then C = 2, F = C+6=8, E = 7-C-G= 4 
Or C = 3, F = 9, E = 3 (invalid,since the same C,E value) 
 
Case(ii) When G =2, D = G+6= 8,  
and then C = 1, F = C+6=7, E= 7-C-G=4 
Or C = 3, F = 9, E = 2(invalid,since the same C,G value) 
 
Case(iii) When G =3, D = G+6=9, 
and then C = 1, F = C+6=7, E =7-C-G= 3(invalid,since the same E,G value) 
Or C = 2, F = C+6=8, E=7-C-G = 2(invalid,since the same C,E value) 
 
In these three possible cases of the values of G, we obtain that the only valid value of E is 4 

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Also, to add, the values of ABCDEFGHI  can be expressed as

(1,3,9), (1,8,4), 2,4,7), (2,5,6)

possible combo for 13 is
9+1+2
2+3+8
3+4+6
2+5+6

some digits are repeated  like 2 , 3 and 6..it means they are to be positioned at the junction
the logical pos comes out to be

9 1 2
      8
      3 4 6
            5
            2

therfore c = 3 from above fig