a maths question~!

For simplicity, ABC will mean area of triangle ABC and so on.
Let AGE = a, so EGB = 2a ( same height and base length ratio = AE : EB = 1 : 2).
Let BHF = b, so FHC = 2b.
Let AID = c, so DIC = 2c.
Let GBH = x, HCI = y and AGI = z. Also, GHI = 4 (given).
AIE : EIB = 1 : 2
(a + z) : (2a + x + 4) = 1 : 2
2(a + z) = 2a + x + 4
2a + 2z = 2a + x + 4
2z = x + 4...........................(1)
Similarly, we get
DHC : AHD = 2 : 1
(2c + y) : (c + z + 4) = 2 : 1
2c + y = 2c + 2z + 8
y = 2z + 8....................(2)
GBF : GFC = 1 : 2
(x + b) : (2b + y + 4) = 1 : 2
2x + 2b = 2b + y + 4
2x = y + 4 .....................(3)
Eliminating 2z from (1) and (2) we get
y = x + 12...........(4)
From (3) and (4)
2x = x + 12 + 4
x = 16, y = x + 12 = 28 and z = (y - 8)/2 = 10.
Applying the same method,
ACE : ECB = 1 : 2
(3c + a + z) : (3b + 2a + x + y + 4) = 1 : 2
2(3c + a + 10) = 3b + 2a + 48
6c + 2a + 20 = 3b + 2a + 48
6c = 3b + 28..................(5)
ABF : AFC = 1 : 2
(3a + b + x) : (3c + 2b + y + z + 4) = 1 : 2
2(3a + b + 16) = 3c + 2b + 42
6a = 3c + 10...............(6)
CBD : DBA = 2 : 1
(2c + 3b + y) : (3a + c + x + z + 4) = 2 : 1
2c + 3b + 28 = 6a + 2c + 60
3b = 6a + 32................(7)
Adding (5), (6) and (7) together we get
6a + 6c + 3b = 3b + 28 + 3c + 10 + 6a + 32
3c = 70
3a = (3c + 10)/2 = 40
3b = 80 + 32 = 112.
So area of triangle ABC = 3a + 3b + 3c + x + y + z + 4
= 112 + 40 + 70 + 16 + 28 + 10 + 4
= 280.