Okay, so I need to find the area of triangle ENG in triangle ABC where the area of ABC is 144. Let me start by understanding the problem step by step.

First, triangle ABC has medians AD, BE, and CF intersecting at the centroid G. The centroid divides each median into a ratio of 2:1, right? So AG:GD = 2:1, BG:GE = 2:1, and CG:GF = 2:1.

Now, there's a line through G that's parallel to BC. This line intersects AB at M and AC at N. So, MN is parallel to BC and passes through G. I need to figure out the coordinates or positions of points M, N, E, and G to find the area of triangle ENG.

Maybe using coordinate geometry would help here. Let me assign coordinates to the triangle ABC. Let's assume ABC is a coordinate system where point A is at (0, 0). Let me place point B at (2b, 0) and point C at (2c, 2d). Hmm, but maybe it's better to simplify. Let's make ABC a triangle with coordinates: A at (0, 0), B at (2, 0), and C at (0, 2). Wait, but then the area would be 2, but the area is given as 144. Maybe scaling will be needed later.

Alternatively, let's use a more general coordinate system. Let me set point A at (0, 0), point B at (2a, 0), and point C at (0, 2c). Then the area of triangle ABC would be (1/2)*base*height = (1/2)*(2a)*(2c) = 2ac. Since the area is 144, then 2ac = 144, so ac = 72. I can keep this in mind.

Now, the centroid G is the intersection of the medians. The coordinates of G can be found by averaging the coordinates of the vertices. So G_x = (0 + 2a + 0)/3 = (2a)/3, G_y = (0 + 0 + 2c)/3 = (2c)/3. So G is at ((2a/3), (2c/3)).

Next, the line through G parallel to BC. Let's find the equation of BC. Points B (2a, 0) and C (0, 2c). The slope of BC is (2c - 0)/(0 - 2a) = -c/a. Therefore, the line MN, which is parallel to BC, will have the same slope, -c/a.

The equation of line MN passing through G ((2a/3), (2c/3)) is:

y - (2c/3) = (-c/a)(x - 2a/3)

Simplify this equation:

y = (-c/a)x + (2c/a)(a/3) + 2c/3

Wait, let me compute that step by step.

The equation is y = (-c/a)(x - 2a/3) + 2c/3

Multiply out:

y = (-c/a)x + (-c/a)*(-2a/3) + 2c/3

Simplify the terms:

(-c/a)*(-2a/3) = (2c/3)

So y = (-c/a)x + 2c/3 + 2c/3 = (-c/a)x + 4c/3

So the equation of MN is y = (-c/a)x + 4c/3

Now, this line intersects AB and AC at points M and N. Let's find coordinates of M and N.

First, find M on AB. AB is from A(0,0) to B(2a,0). The equation of AB is y = 0. So set y = 0 in the equation of MN:

0 = (-c/a)x + 4c/3

Solve for x:

(-c/a)x = -4c/3

Multiply both sides by (-a/c):

x = (-4c/3)*(-a/c) = (4a)/3

So point M is at (4a/3, 0)

Wait, but AB is from (0,0) to (2a,0). So x ranges from 0 to 2a. But 4a/3 is less than 2a (since 4a/3 ≈ 1.333a and 2a is 2a). So that's okay. So M is at (4a/3, 0)

Similarly, find point N on AC. AC is from A(0,0) to C(0,2c). The equation of AC is x = 0. Wait, but MN is supposed to intersect AC. Wait, but AC is the line x=0. Let's check if the line MN intersects AC at x=0.

Substitute x=0 into the equation of MN:

y = (-c/a)(0) + 4c/3 = 4c/3

So point N is at (0, 4c/3)

But AC is from (0,0) to (0,2c). So 4c/3 is less than 2c (since 4c/3 ≈ 1.333c and 2c is 2c). So N is on AC.

Therefore, points M(4a/3, 0) and N(0, 4c/3). So the line MN connects (4a/3, 0) to (0, 4c/3). The length of MN can be calculated, but maybe that's not necessary.

Now, we need to find the area of triangle ENG. So points E, G, and N.

First, let's find the coordinates of E. E is the midpoint of AC, since BE is a median. Since AC is from A(0,0) to C(0,2c), the midpoint E is at (0, c)

Wait, but BE is a median from B to AC. So E is the midpoint of AC. So E is at ( (0+0)/2, (0 + 2c)/2 ) = (0, c). Correct.

So E is at (0, c)

G is at (2a/3, 2c/3)

N is at (0, 4c/3)

Therefore, triangle ENG has vertices at E(0, c), G(2a/3, 2c/3), and N(0, 4c/3)

Wait, let's plot these points. E is at (0, c), G is at (2a/3, 2c/3), and N is at (0, 4c/3). So triangle ENG is formed by these three points.

Wait, E and N are both on the y-axis at x=0, so the base of the triangle could be from E(0,c) to N(0,4c/3). The length of EN is 4c/3 - c = c/3. Then the height would be the x-coordinate of G, which is 2a/3, since the base is vertical.

But actually, triangle ENG is a triangle with vertices at (0,c), (2a/3, 2c/3), and (0,4c/3). Let me use the formula for the area of a triangle given coordinates.

The area can be calculated using the determinant formula:

Area = (1/2)| (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) |

Plugging in the coordinates:

x1 = 0, y1 = c

x2 = 2a/3, y2 = 2c/3

x3 = 0, y3 = 4c/3

So:

Area = (1/2)| 0*(2c/3 - 4c/3) + (2a/3)*(4c/3 - c) + 0*(c - 2c/3) |

Simplify each term:

First term: 0*(something) = 0

Second term: (2a/3)*(4c/3 - 3c/3) = (2a/3)*(c/3) = (2a c)/9

Third term: 0*(something) = 0

So Area = (1/2)| 0 + (2ac)/9 + 0 | = (1/2)*(2ac/9) = ac/9

But earlier, we found that ac = 72 (since area of ABC is 144 = 2ac, so ac = 72)

Therefore, area of triangle ENG is 72/9 = 8.

Wait, so the area is 8? Let me check if that makes sense.

Alternatively, maybe using vectors or ratios would confirm this. Let's think differently.

Since MN is parallel to BC and passes through centroid G, which divides the centroid in a 2:1 ratio.

In a triangle, a line drawn through the centroid parallel to a side will divide the other two sides in a 2:1 ratio. Wait, but here MN is parallel to BC and passes through G. So in triangle ABC, the line MN is parallel to BC and passes through G. Then, the ratio of division on AB and AC should be the same.

Since the centroid divides the median in a 2:1 ratio, and MN is parallel to BC, the line MN is at a height from BC. Wait, maybe using similarity.

Since MN is parallel to BC, triangle AMN is similar to triangle ABC. The ratio of similarity would be determined by the distance from G to BC.

But the centroid divides the median AD in a 2:1 ratio. The distance from A to G is 2/3 of AD, and from G to BC is 1/3 of AD. Since MN passes through G and is parallel to BC, the distance between MN and BC is 1/3 of the height of the triangle.

Therefore, the ratio of similarity between triangle AMN and ABC is (height of AMN)/(height of ABC) = (height - 1/3 height)/height = 2/3. Wait, but the line MN is closer to BC, so the height from MN to BC is 1/3 the height of ABC, so the height from A to MN is 2/3 the height of ABC. Therefore, the ratio of similarity is 2/3. Therefore, the sides of triangle AMN are 2/3 the sides of ABC. So the area of triangle AMN would be (2/3)^2 * 144 = (4/9)*144 = 64.

But triangle AMN has area 64, and triangle ABC has area 144. Then the area between AMN and ABC is 144 - 64 = 80. But how does this relate to triangle ENG?

Alternatively, maybe the coordinates approach was correct. The area came out to 8. Let me check again.

Wait, in the coordinate system, the area of triangle ABC is 144, and ac = 72. Then triangle ENG has area ac/9 = 72/9 = 8. So that's 8. But let me confirm with another method.

Alternatively, coordinates:

Points E(0, c), G(2a/3, 2c/3), N(0, 4c/3)

Compute vectors EG and EN.

Vector EG = G - E = (2a/3 - 0, 2c/3 - c) = (2a/3, -c/3)

Vector EN = N - E = (0 - 0, 4c/3 - c) = (0, c/3)

The area is 1/2 |EG x EN| = 1/2 | (2a/3)(c/3) - (-c/3)(0) | = 1/2 | (2ac)/9 | = ac/9 = 72/9 = 8. Same result.

Therefore, the area is 8. So the final answer is \boxed{8}

**Final Answer**
\boxed{8}
Thus, the area of triangle $ENG$ is $\boxed{8}$.
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