Explanation:
Okay, so I need to find the area of triangle ENG in triangle ABC, where the medians AD, BE, and CF intersect at centroid G. There's a line through G parallel to BC that intersects AB and AC at M and N, respectively. The area of triangle ABC is given as 144. Hmm, let me try to visualize this first.

First, let me recall that the centroid divides each median into a ratio of 2:1, with the longer part being closer to the vertex. So, for example, AG:GD = 2:1, BG:GE = 2:1, and CG:GF = 2:1. That might come in handy later.

Now, there's a line through G that's parallel to BC. Let me try to sketch this mentally. Since BC is the base of the triangle, a line parallel to BC through G would be somewhere inside the triangle. This line intersects AB at M and AC at N. So, MN is parallel to BC and passes through centroid G. I need to find the area of triangle ENG. Wait, who is E? Let me check the problem again.

Wait, the problem mentions medians AD, BE, and CF. So, BE is a median from B to AC. Therefore, E is the midpoint of AC. Similarly, F is the midpoint of AB, and D is the midpoint of BC. So, E is the midpoint of AC. Therefore, triangle ENG is formed by points E, N, and G. So, E is midpoint of AC, N is the intersection of the line through G parallel to BC with AC, and G is the centroid. Wait, but the line through G parallel to BC intersects AB at M and AC at N. So, N is on AC, and M is on AB. So, triangle ENG is formed by E (midpoint of AC), N (some point on AC), and G (centroid). Wait, but if N is on AC, and E is also on AC, then triangle ENG would have two points on AC and one at G. So, that would make triangle ENG a triangle with base EN on AC and vertex at G. But maybe I need to confirm the exact positions.

Alternatively, maybe there's a typo? Wait, the problem says "the line through G that is parallel to BC intersects AB and AC at M and N, respectively." So, the line passes through G, is parallel to BC, and intersects AB at M and AC at N. Therefore, MN is parallel to BC, passing through G. So, points M and N are on AB and AC respectively. Then, triangle ENG is formed by E (midpoint of AC), N (on AC), and G (centroid). Wait, but if N is on AC, and E is also on AC, then the triangle ENG would have two points on AC. So, the base is EN, and the height would be the distance from G to AC? But maybe I need to find coordinates to figure this out.

Alternatively, maybe I can use coordinate geometry. Let me assign coordinates to triangle ABC. Let me choose coordinates such that calculations are easier. Let me set coordinates with point A at (0, 0), B at (2b, 0), and C at (2c, 2d). Wait, but maybe even simpler. Since medians are involved, maybe using coordinates where centroid is at (0, 0)? Wait, but maybe it's better to use standard coordinates.

Alternatively, let me take coordinates where BC is horizontal for simplicity. Wait, but since the line through G is parallel to BC, maybe setting BC horizontal would make things easier. Let me try that.

Let me set coordinate system with point B at (0, 0), point C at (2c, 0), so that BC is along the x-axis. Then, since the centroid G divides the medians in ratio 2:1, but maybe I need coordinates for A. Let me denote coordinates for A as (a, b). Then, coordinates of centroid G would be the average of the coordinates of A, B, and C. So, G_x = (a + 0 + 2c)/3, G_y = (b + 0 + 0)/3 = b/3.

Now, the line through G parallel to BC. Since BC is horizontal (from (0,0) to (2c, 0)), so BC is horizontal, so the line through G parallel to BC is also horizontal. Therefore, the line through G has equation y = G_y = b/3. This line intersects AB and AC at points M and N.

So, let me find coordinates of M and N. First, find equations of AB and AC.

Point A is (a, b), point B is (0, 0). The equation of AB can be parametrized or expressed as a line. Let me find the equation of AB. The slope of AB is (0 - b)/(0 - a) = -b / (-a) = b/a. Therefore, equation of AB is y = (b/a)x.

Similarly, equation of AC: point A (a, b) and point C (2c, 0). The slope is (0 - b)/(2c - a) = -b/(2c - a). Therefore, equation of AC is y - b = [-b/(2c - a)](x - a).

Now, the horizontal line through G is y = b/3. Let me find point M on AB where y = b/3. So, substituting y = b/3 into AB's equation:

b/3 = (b/a)x => x = (b/3) * (a/b) = a/3. Therefore, coordinates of M are (a/3, b/3).

Similarly, find point N on AC where y = b/3. Let me substitute y = b/3 into equation of AC:

b/3 - b = [-b/(2c - a)](x - a)

=> -2b/3 = [-b/(2c - a)](x - a)

Multiply both sides by (2c - a)/(-b):

(-2b/3) * [(2c - a)/(-b)] = x - a

Simplify:

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

Therefore, x = a + (2/3)(2c - a) = a + (4c/3 - 2a/3) = (a - 2a/3) + 4c/3 = (a/3) + 4c/3 = (a + 4c)/3

Therefore, coordinates of N are ((a + 4c)/3, b/3).

So, points M (a/3, b/3) and N ((a + 4c)/3, b/3). Now, we need to find area of triangle ENG. First, let me recall that E is the midpoint of AC. Coordinates of E: midpoint of A(a, b) and C(2c, 0) is ((a + 2c)/2, (b + 0)/2) = ((a + 2c)/2, b/2).

So, coordinates:

E: ((a + 2c)/2, b/2)

N: ((a + 4c)/3, b/3)

G: ((a + 0 + 2c)/3, b/3) = ((a + 2c)/3, b/3)

Wait, so G is ((a + 2c)/3, b/3). Let me check that again. Earlier, centroid coordinates are ((a + 0 + 2c)/3, (b + 0 + 0)/3) = ((a + 2c)/3, b/3). Yes, correct.

So, coordinates of E, N, G:

E: ((a + 2c)/2, b/2)

N: ((a + 4c)/3, b/3)

G: ((a + 2c)/3, b/3)

So, triangle ENG has coordinates E, N, G. Let me write them again:

E: ( (a + 2c)/2 , b/2 )

N: ( (a + 4c)/3 , b/3 )

G: ( (a + 2c)/3 , b/3 )

To find the area of triangle ENG, I can use the shoelace formula. Let me denote coordinates as:

E: (x1, y1) = ((a + 2c)/2, b/2)

N: (x2, y2) = ((a + 4c)/3, b/3)

G: (x3, y3) = ((a + 2c)/3, b/3)

Shoelace formula:

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

Let me compute each term.

First, compute y2 - y3: b/3 - b/3 = 0. So, first term is x1 * 0 = 0.

Second term: x2(y3 - y1) = x2*(b/3 - b/2) = x2*(-b/6)

Third term: x3(y1 - y2) = x3*(b/2 - b/3) = x3*(b/6)

Therefore, Area = | [0 + x2*(-b/6) + x3*(b/6)] / 2 | = | [ (-x2 + x3) * b / 6 ] / 2 | = | [ (-x2 + x3) * b ] / 12 |.

So, compute (-x2 + x3):

x2 = (a + 4c)/3, x3 = (a + 2c)/3. Therefore, -x2 + x3 = - (a + 4c)/3 + (a + 2c)/3 = [ -a -4c + a + 2c ] / 3 = (-2c)/3.

Therefore, Area = | [ (-2c)/3 * b ] / 12 | = | (-2bc)/36 | = | -bc/18 | = bc/18.

So, the area of triangle ENG is bc/18. Now, we need to relate this to the area of triangle ABC, which is given as 144.

Let me compute area of ABC. Coordinates of A(a, b), B(0, 0), C(2c, 0). Using shoelace formula:

Area ABC = | (a*(0 - 0) + 0*(0 - b) + 2c*(b - 0))/2 | = | 0 + 0 + 2c*b / 2 | = | cb | = | bc |. Since area is positive, it's bc. Given that area ABC is 144, so bc = 144.

Therefore, area of triangle ENG is bc/18 = 144 / 18 = 8. So, the area is 8? Wait, that seems straightforward. But let me check my steps again to make sure.

First, coordinates assigned: A(a, b), B(0, 0), C(2c, 0). Then centroid G is ((a + 2c)/3, b/3). Then line through G parallel to BC (which is horizontal) is y = b/3. Found points M and N by intersecting this line with AB and AC. Then found coordinates of E as midpoint of AC. Then applied shoelace formula for triangle ENG. Calculated the area as bc/18, and since area ABC is bc = 144, then bc/18 = 8. So, answer is 8? Wait, but let me check if there's a mistake in coordinates.

Wait, area of ABC: with coordinates A(a, b), B(0, 0), C(2c, 0). The area can also be calculated as base * height / 2. The base BC is 2c units long (from 0 to 2c on x-axis). The height is the y-coordinate of A, which is b. Therefore, area ABC is (2c * b)/2 = bc. Which matches. So bc = 144. Therefore, bc/18 = 144 / 18 = 8. So, area of triangle ENG is 8. Therefore, answer is 8? But let me check with another approach to confirm.

Alternatively, maybe using ratios and properties of centroids and similar triangles.

Since MN is parallel to BC and passes through centroid G. Let me recall that in a triangle, a line parallel to the base divides the sides proportionally. Also, since G is centroid, which divides the medians in ratio 2:1. Let me think about the position of MN.

Since MN is parallel to BC and passes through centroid G. Let me consider the ratio of distances from the base. The centroid is located at 1/3 of the height from the base. Wait, if we consider the height from A to BC, then centroid is located at 1/3 of that height. Therefore, if MN is parallel to BC and passes through centroid, then the distance from MN to BC is 1/3 of the height? Wait, no. Wait, if the centroid is located at 1/3 of the height from the base BC, then the line MN passing through centroid would be located at 1/3 of the height from BC. Therefore, the ratio of lengths of MN to BC would be?

Since the line is parallel to BC, the ratio of lengths is equal to the ratio of distances from A. Wait, if the distance from A to MN is 2/3 of the total height, then MN would be 2/3 of BC? Wait, no. Wait, if the distance from MN to BC is 1/3 of the height, then the distance from A to MN is 2/3 of the height. Therefore, the ratio of similarity is 2/3, so MN = (2/3) BC. Wait, but MN is passing through centroid. Wait, but let me think again.

Let me denote h as the height from A to BC. Then centroid G is located at h/3 from BC. Therefore, the line MN is at height h/3 from BC. Therefore, the distance from A to MN is h - h/3 = 2h/3. Therefore, the ratio of similarity between triangle AMN and triangle ABC is 2/3. Therefore, area of AMN is (2/3)^2 * area ABC = 4/9 * 144 = 64. Therefore, area of AMN is 64. Then, area of ABC is 144, so area of trapezoid MNBC is 144 - 64 = 80. But how does this help with triangle ENG?

Alternatively, maybe I need to find coordinates of E, N, G again. Wait, but earlier calculation gave area of ENG as 8. Let me check with this approach.

Alternatively, since E is midpoint of AC, and N is a point on AC. Let me find the ratio AN / AC. Since MN is parallel to BC, and passes through centroid G. Let me find where N is located on AC.

Earlier, in coordinate terms, N was at ((a + 4c)/3, b/3). Coordinates of A(a, b), C(2c, 0). Let me compute the ratio AN / AC.

Parametrize AC. Let me consider vector from A to C: (2c - a, -b). The coordinates of N are ((a + 4c)/3, b/3). Let me express N as A + t*(vector AC). So:

x-coordinate: a + t*(2c - a) = (a + 4c)/3

Similarly, y-coordinate: b + t*(-b) = b/3

From y-coordinate: b - tb = b/3 => 1 - t = 1/3 => t = 2/3.

Therefore, t = 2/3. Therefore, AN / AC = 2/3. Therefore, N divides AC in ratio AN:NC = 2:1. Wait, but E is the midpoint of AC, so AE:EC = 1:1. Therefore, E is located at 1/2 of AC, and N is located at 2/3 of AC. Therefore, the distance from E to N is (2/3 - 1/2) = 1/6 of AC. Therefore, EN = (1/6) AC.

But how does this help? Maybe we can find the area of ENG. Since G is centroid, and we know coordinates or ratios.

Alternatively, since we know coordinates, maybe we can find vectors or use area ratios.

Alternatively, since we found earlier that area of ENG is bc/18, and bc = 144, so 144/18 = 8. So, answer is 8. But let me check with another approach.

Let me consider mass point or area ratios.

Since MN is parallel to BC and passes through centroid G. As earlier, the ratio of AN / AC = 2/3. Therefore, N divides AC into AN:NC = 2:1. Similarly, M divides AB into AM:MB = 2:1? Wait, earlier in coordinates, M was at (a/3, b/3). Since AB goes from A(a, b) to B(0, 0). So, coordinates of M are (a/3, b/3). Therefore, vector from A to M is (a/3 - a, b/3 - b) = (-2a/3, -2b/3). Therefore, AM is 1/3 of AB? Wait, no. Wait, if you go from A to B, the coordinates of M are (a/3, b/3). So, the ratio AM / AB. Let me parametrize AB. Let me say AB is from A(a, b) to B(0, 0). Let me write parametric equations. Let parameter t go from 0 to 1, with t=0 at A and t=1 at B. Then, coordinates are (a - a*t, b - b*t). So, when does this equal (a/3, b/3)? Let me solve:

a - a*t = a/3 => 1 - t = 1/3 => t = 2/3. Similarly for y-coordinate. Therefore, t = 2/3. Therefore, AM / AB = 2/3. Therefore, AM = 2/3 AB, so MB = 1/3 AB. Therefore, M divides AB in ratio AM:MB = 2:1. Similarly, N divides AC in ratio AN:NC = 2:1.

Therefore, points M and N are located 2/3 along AB and AC respectively from A. Therefore, triangle AMN is similar to triangle ABC with ratio 2/3, so area 4/9 of ABC, which is 4/9 * 144 = 64, as before.

Now, E is the midpoint of AC. So, AE = 1/2 AC. Since AN = 2/3 AC, then EN = AN - AE = 2/3 AC - 1/2 AC = (4/6 - 3/6) AC = 1/6 AC. Therefore, EN = 1/6 AC. So, the length of EN is 1/6 of AC.

But how does this relate to the area of triangle ENG?

Let me think. Since G is centroid, coordinates of G are known. Alternatively, maybe we can find the area of ENG by considering coordinates or ratios.

Alternatively, since we know coordinates of E, N, G, and we found area as 8, which is 144 / 18. Let me check if 18 is related to the ratios. Since area of ABC is 144, and area of ENG is 8, which is 144 / 18. Maybe there's a way to see this ratio.

Alternatively, since EN is 1/6 of AC, and the height from G to EN? Wait, but EN is on AC. Wait, no. Wait, E and N are both on AC, so EN is a segment on AC. Then, triangle ENG has base EN on AC and vertex at G. Therefore, the area of triangle ENG is (1/2)*EN*height, where height is the distance from G to AC.

Wait, but since E and N are on AC, the base is EN, and the height would be the distance from G to line AC. Wait, but if I can compute that distance, then I can compute the area.

Alternatively, since we know coordinates, maybe compute the distance from G to AC.

But let me try this approach. Let me compute the distance from G to AC. Then, area of triangle ENG is 1/2 * EN * distance.

First, find EN. As above, EN = 1/6 AC. The length of AC can be found from coordinates. Coordinates of A(a, b) and C(2c, 0). So, AC has length sqrt[(2c - a)^2 + (0 - b)^2]. But maybe we don't need the actual length, since we can relate areas.

Alternatively, since area of triangle ABC is 144, which is 1/2 * BC * height from A. But maybe we need to relate the distance from G to AC.

Alternatively, since centroid divides the medians in ratio 2:1, the distance from centroid to AC would be 1/3 of the distance from B to AC? Wait, no. Wait, the distance from centroid to AC. Let me think.

The centroid is located 1/3 of the distance from the midpoint of the median. Wait, maybe it's easier to compute using coordinates.

Coordinates of G: ((a + 2c)/3, b/3). Equation of AC: we had earlier y - b = [-b/(2c - a)](x - a). Let me write it in standard form.

Let me rearrange the equation of AC:

y = [-b/(2c - a)](x - a) + b

Multiply both sides by (2c - a):

(2c - a)y = -b(x - a) + b(2c - a)

=> (2c - a)y = -bx + ab + 2bc - ab

=> (2c - a)y = -bx + 2bc

Bring all terms to left:

bx + (2c - a)y - 2bc = 0

So, equation of AC is bx + (2c - a)y - 2bc = 0.

Distance from point G((a + 2c)/3, b/3) to line AC is:

|b*( (a + 2c)/3 ) + (2c - a)*(b/3) - 2bc| / sqrt(b² + (2c - a)²)

Let me compute numerator:

First term: b*(a + 2c)/3

Second term: (2c - a)*(b)/3

Third term: -2bc

So, total numerator:

[ b(a + 2c)/3 + b(2c - a)/3 - 2bc ] = [ b(a + 2c + 2c - a)/3 - 2bc ] = [ b(4c)/3 - 2bc ] = (4bc/3 - 6bc/3) = (-2bc)/3

Absolute value: | -2bc/3 | = 2bc/3

Therefore, distance from G to AC is (2bc/3) / sqrt(b² + (2c - a)^2 )

But the area of triangle ENG is 1/2 * EN * distance.

Earlier, we found EN = 1/6 AC. The length of AC is sqrt[(2c - a)^2 + b²]. Therefore, EN = (1/6) * sqrt[(2c - a)^2 + b²]

Therefore, area of ENG is 1/2 * (1/6) * sqrt(...) * [ (2bc/3) / sqrt(...) ] = 1/2 * 1/6 * 2bc/3 = (1/6)*(bc/3) = bc/18, which matches our previous result. Therefore, area of ENG is bc/18 = 144 / 18 = 8. Therefore, answer is 8.

But let me check once more with another approach. Let me use vectors.

Let me take coordinates again. Let me assume coordinates for simplicity. Let me choose specific coordinates for triangle ABC to make calculations easier. Let me set coordinates such that:

Let me take A(0, 0), B(0, 0)... Wait, no. Let me choose coordinates where calculations are easier. Let me take A(0, 0), B(2, 0), C(0, 2). Wait, but then area of ABC is 2, but we need area 144. Maybe scale later. Let me try.

Let me take A(0, 0), B(2, 0), C(0, 2). Then, area of ABC is 2. Let me compute coordinates of centroid G: ((0 + 2 + 0)/3, (0 + 0 + 2)/3) = (2/3, 2/3). Then, line through G parallel to BC. BC is from (2, 0) to (0, 2), so slope is (2 - 0)/(0 - 2) = -1. Wait, but in this case, BC is not horizontal. Wait, earlier I considered BC horizontal, but here BC has slope -1. Maybe this complicates things. Alternatively, let me take BC horizontal.

Let me take A(0, h), B(-1, 0), C(1, 0). Then, BC is from (-1, 0) to (1, 0), so length 2, and area of ABC is (base * height)/2 = (2 * h)/2 = h. Let me set h = 144, but maybe take h = 1 for simplicity, then scale later. Wait, but maybe this is getting too convoluted.

Alternatively, take A(0, 3), B(0, 0), C(2, 0). Then, area of ABC is (base * height)/2 = (2 * 3)/2 = 3. Then centroid G is ((0 + 0 + 2)/3, (3 + 0 + 0)/3) = (2/3, 1). Then, line through G parallel to BC. Since BC is horizontal (from (0,0) to (2, 0)), so line through G is horizontal line y = 1. This intersects AB and AC.

Equation of AB: from (0, 3) to (0, 0). Wait, AB is vertical line x = 0. Wait, but line y = 1 intersects AB at (0, 1). Similarly, equation of AC: from (0, 3) to (2, 0). Slope is (0 - 3)/(2 - 0) = -3/2. Equation: y - 3 = -3/2 (x - 0) => y = -3/2 x + 3. Intersection with y = 1:

1 = -3/2 x + 3 => -3/2 x = -2 => x = (-2) * (-2/3) = 4/3. Therefore, point N is (4/3, 1). So, points M(0, 1) and N(4/3, 1). E is midpoint of AC: coordinates of A(0, 3) and C(2, 0), so midpoint E is ((0 + 2)/2, (3 + 0)/2) = (1, 1.5). G is (2/3, 1). So, triangle ENG has coordinates E(1, 1.5), N(4/3, 1), G(2/3, 1). Let me compute its area.

Using shoelace formula:

Coordinates:

E(1, 1.5), N(4/3, 1), G(2/3, 1)

Compute area:

Arrange points:

E(1, 3/2), N(4/3, 1), G(2/3, 1)

Compute using formula:

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

Plugging in:

x1 = 1, y1 = 3/2

x2 = 4/3, y2 = 1

x3 = 2/3, y3 = 1

Compute:

1*(1 - 1) + (4/3)*(1 - 3/2) + (2/3)*(3/2 - 1)

= 1*(0) + (4/3)*(-1/2) + (2/3)*(1/2)

= 0 - 4/6 + 2/6 = (-4 + 2)/6 = -2/6 = -1/3

Take absolute value and divide by 2? Wait, no. Wait, the formula is |sum / 2|. Wait, the formula is |sum| / 2. Wait, the formula is:

Area = | [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] / 2 |.

So, the sum inside is -1/3, so area is | -1/3 / 2 |? Wait no, the entire expression is divided by 2. Wait, no:

Wait, the formula is:

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

So, the sum inside is -1/3, so area is | (-1/3) / 2 |? Wait, no. Wait, the entire expression is divided by 2. Wait, no, the formula is:

It's [sum] divided by 2. So, the sum is -1/3, so area is | -1/3 / 2 |? Wait, no, no. Wait, the formula is:

Area = | [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] / 2 |.

So, the numerator is -1/3, so area is | (-1/3) / 2 |? No, no. Wait, the numerator is already calculated as -1/3. Wait, no:

Wait, let me recalculate:

Original calculation:

x1(y2 - y3) = 1*(1 - 1) = 0

x2(y3 - y1) = (4/3)*(1 - 3/2) = (4/3)*(-1/2) = -2/3

x3(y1 - y2) = (2/3)*(3/2 - 1) = (2/3)*(1/2) = 1/3

Therefore, total sum is 0 - 2/3 + 1/3 = -1/3. Then divide by 2: (-1/3)/2 = -1/6. Take absolute value: 1/6. Therefore, area is 1/6.

But in this case, area of ABC is 3, and area of ENG is 1/6. Therefore, ratio is 1/6 divided by 3 = 1/18. Therefore, area of ENG is (1/18)*area ABC. Therefore, if area ABC is 144, then area ENG is 144 / 18 = 8. Which matches our previous result. Therefore, this confirms that the area is indeed 8. Therefore, the answer is 8.

Therefore, I think the answer is \boxed{8}.

**Final Answer**
\boxed{8}

To solve the problem, we are given a triangle $ ABC $ with medians $ AD $, $ BE $, and $ CF $ intersecting at the centroid $ G $. A line through $ G $, parallel to $ BC $, intersects $ AB $ at $ M $ and $ AC $ at $ N $. We are to find the area of triangle $ ENG $, where $ E $ is the midpoint of $ AC $, and the area of triangle $ ABC $ is given as 144.

---

### Step 1: Coordinate Setup

To simplify the geometry, we place triangle $ ABC $ in the coordinate plane:

- Let $ A = (0, 0) $
- Let $ B = (2b, 0) $
- Let $ C = (2c, 2d) $

This setup ensures that the base $ BC $ lies along the x-axis, and the triangle is symmetric in a way that simplifies calculations.

---

### Step 2: Centroid Coordinates

The centroid $ G $ of triangle $ ABC $ is the average of the coordinates of the vertices:

$$
G = \left( \frac{0 + 2b + 2c}{3}, \frac{0 + 0 + 2d}{3} \right) = \left( \frac{2b + 2c}{3}, \frac{2d}{3} \right)
$$

---

### Step 3: Line Through $ G $ Parallel to $ BC $

Since $ BC $ lies along the x-axis, the line through $ G $ that is parallel to $ BC $ is horizontal. Therefore, its equation is:

$$
y = \frac{2d}{3}
$$

This line intersects:

- **Line $ AB $** at point $ M $
- **Line $ AC $** at point $ N $

---

### Step 4: Coordinates of Points $ M $, $ N $, and $ E $

#### Point $ M $ on $ AB $:

Line $ AB $ goes from $ A = (0, 0) $ to $ B = (2b, 0) $, so it's horizontal. The line $ y = \frac{2d}{3} $ intersects $ AB $ at:

$$
M = \left( \frac{2b}{3}, \frac{2d}{3} \right)
$$

#### Point $ N $ on $ AC $:

Line $ AC $ goes from $ A = (0, 0) $ to $ C = (2c, 2d) $. The slope of $ AC $ is:

$$
m = \frac{2d - 0}{2c - 0} = \frac{d}{c}
$$

So the equation of line $ AC $ is:

$$
y = \frac{d}{c}x
$$

Setting $ y = \frac{2d}{3} $, we solve for $ x $:

$$
\frac{2d}{3} = \frac{d}{c}x \Rightarrow x = \frac{2c}{3}
$$

So, point $ N = \left( \frac{2c}{3}, \frac{2d}{3} \right) $

#### Point $ E $, the midpoint of $ AC $:

$$
E = \left( \frac{0 + 2c}{2}, \frac{0 + 2d}{2} \right) = \left( c, d \right)
$$

---

### Step 5: Area of Triangle $ ENG $

We now have the coordinates of the three points:

- $ E = (c, d) $
- $ N = \left( \frac{2c}{3}, \frac{2d}{3} \right) $
- $ G = \left( \frac{2b + 2c}{3}, \frac{2d}{3} \right) $

We apply the **shoelace formula** for the area of triangle $ ENG $:

$$
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
$$

Substitute the coordinates:

- $ x_1 = c, y_1 = d $
- $ x_2 = \frac{2c}{3}, y_2 = \frac{2d}{3} $
- $ x_3 = \frac{2b + 2c}{3}, y_3 = \frac{2d}{3} $

Compute each term:

- $ x_1(y_2 - y_3) = c \left( \frac{2d}{3} - \frac{2d}{3} \right) = 0 $
- $ x_2(y_3 - y_1) = \frac{2c}{3} \left( \frac{2d}{3} - d \right) = \frac{2c}{3} \left( -\frac{d}{3} \right) = -\frac{2cd}{9} $
- $ x_3(y_1 - y_2) = \frac{2b + 2c}{3} \left( d - \frac{2d}{3} \right) = \frac{2b + 2c}{3} \cdot \frac{d}{3} = \frac{(2b + 2c)d}{9} $

Add them:

$$
\text{Area} = \frac{1}{2} \left| 0 - \frac{2cd}{9} + \frac{(2b + 2c)d}{9} \right| = \frac{1}{2} \left| \frac{2bd + 2cd - 2cd}{9} \right| = \frac{1}{2} \cdot \frac{2bd}{9} = \frac{bd}{9}
$$

---

### Step 6: Relating to Area of Triangle $ ABC $

The area of triangle $ ABC $ is:

$$
\text{Area}_{ABC} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2b \cdot 2d = 2bd
$$

Given that $ \text{Area}_{ABC} = 144 $, we have:

$$
2bd = 144 \Rightarrow bd = 72
$$

Now, the area of triangle $ ENG $ is:

$$
\text{Area}_{ENG} = \frac{bd}{9} = \frac{72}{9} = 8
$$

---

### Final Answer

$$
\boxed{8}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of triangle ENG in triangle ABC, where the medians AD, BE, and CF intersect at centroid G. There's a line through G parallel to BC that intersects AB and AC at M and N, respectively. The area of triangle ABC is given as 144. Hmm, let me try to visualize this first.

First, let me recall that the centroid divides each median into a ratio of 2:1, with the longer part being closer to the vertex. So, for example, AG:GD = 2:1, BG:GE = 2:1, and CG:GF = 2:1. That might come in handy later.

Now, there's a line through G that's parallel to BC. Let me try to sketch this mentally. Since BC is the base of the triangle, a line parallel to BC through G would be somewhere inside the triangle. This line intersects AB at M and AC at N. So, MN is parallel to BC and passes through centroid G. I need to find the area of triangle ENG. Wait, who is E? Let me check the problem again.

Wait, the problem mentions medians AD, BE, and CF. So, BE is a median from B to AC. Therefore, E is the midpoint of AC. Similarly, F is the midpoint of AB, and D is the midpoint of BC. So, E is the midpoint of AC. Therefore, triangle ENG is formed by points E, N, and G. So, E is midpoint of AC, N is the intersection of the line through G parallel to BC with AC, and G is the centroid. Wait, but the line through G parallel to BC intersects AB at M and AC at N. So, N is on AC, and M is on AB. So, triangle ENG is formed by E (midpoint of AC), N (some point on AC), and G (centroid). Wait, but if N is on AC, and E is also on AC, then triangle ENG would have two points on AC and one at G. So, that would make triangle ENG a triangle with base EN on AC and vertex at G. But maybe I need to confirm the exact positions.

Alternatively, maybe there's a typo? Wait, the problem says "the line through G that is parallel to BC intersects AB and AC at M and N, respectively." So, the line passes through G, is parallel to BC, and intersects AB at M and AC at N. Therefore, MN is parallel to BC, passing through G. So, points M and N are on AB and AC respectively. Then, triangle ENG is formed by E (midpoint of AC), N (on AC), and G (centroid). Wait, but if N is on AC, and E is also on AC, then the triangle ENG would have two points on AC. So, the base is EN, and the height would be the distance from G to AC? But maybe I need to find coordinates to figure this out.

Alternatively, maybe I can use coordinate geometry. Let me assign coordinates to triangle ABC. Let me choose coordinates such that calculations are easier. Let me set coordinates with point A at (0, 0), B at (2b, 0), and C at (2c, 2d). Wait, but maybe even simpler. Since medians are involved, maybe using coordinates where centroid is at (0, 0)? Wait, but maybe it's better to use standard coordinates.

Alternatively, let me take coordinates where BC is horizontal for simplicity. Wait, but since the line through G is parallel to BC, maybe setting BC horizontal would make things easier. Let me try that.

Let me set coordinate system with point B at (0, 0), point C at (2c, 0), so that BC is along the x-axis. Then, since the centroid G divides the medians in ratio 2:1, but maybe I need coordinates for A. Let me denote coordinates for A as (a, b). Then, coordinates of centroid G would be the average of the coordinates of A, B, and C. So, G_x = (a + 0 + 2c)/3, G_y = (b + 0 + 0)/3 = b/3.

Now, the line through G parallel to BC. Since BC is horizontal (from (0,0) to (2c, 0)), so BC is horizontal, so the line through G parallel to BC is also horizontal. Therefore, the line through G has equation y = G_y = b/3. This line intersects AB and AC at points M and N.

So, let me find coordinates of M and N. First, find equations of AB and AC.

Point A is (a, b), point B is (0, 0). The equation of AB can be parametrized or expressed as a line. Let me find the equation of AB. The slope of AB is (0 - b)/(0 - a) = -b / (-a) = b/a. Therefore, equation of AB is y = (b/a)x.

Similarly, equation of AC: point A (a, b) and point C (2c, 0). The slope is (0 - b)/(2c - a) = -b/(2c - a). Therefore, equation of AC is y - b = [-b/(2c - a)](x - a).

Now, the horizontal line through G is y = b/3. Let me find point M on AB where y = b/3. So, substituting y = b/3 into AB's equation:

b/3 = (b/a)x => x = (b/3) * (a/b) = a/3. Therefore, coordinates of M are (a/3, b/3).

Similarly, find point N on AC where y = b/3. Let me substitute y = b/3 into equation of AC:

b/3 - b = [-b/(2c - a)](x - a)

=> -2b/3 = [-b/(2c - a)](x - a)

Multiply both sides by (2c - a)/(-b):

(-2b/3) * [(2c - a)/(-b)] = x - a

Simplify:

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

Therefore, x = a + (2/3)(2c - a) = a + (4c/3 - 2a/3) = (a - 2a/3) + 4c/3 = (a/3) + 4c/3 = (a + 4c)/3

Therefore, coordinates of N are ((a + 4c)/3, b/3).

So, points M (a/3, b/3) and N ((a + 4c)/3, b/3). Now, we need to find area of triangle ENG. First, let me recall that E is the midpoint of AC. Coordinates of E: midpoint of A(a, b) and C(2c, 0) is ((a + 2c)/2, (b + 0)/2) = ((a + 2c)/2, b/2).

So, coordinates:

E: ((a + 2c)/2, b/2)

N: ((a + 4c)/3, b/3)

G: ((a + 0 + 2c)/3, b/3) = ((a + 2c)/3, b/3)

Wait, so G is ((a + 2c)/3, b/3). Let me check that again. Earlier, centroid coordinates are ((a + 0 + 2c)/3, (b + 0 + 0)/3) = ((a + 2c)/3, b/3). Yes, correct.

So, coordinates of E, N, G:

E: ((a + 2c)/2, b/2)

N: ((a + 4c)/3, b/3)

G: ((a + 2c)/3, b/3)

So, triangle ENG has coordinates E, N, G. Let me write them again:

E: ( (a + 2c)/2 , b/2 )

N: ( (a + 4c)/3 , b/3 )

G: ( (a + 2c)/3 , b/3 )

To find the area of triangle ENG, I can use the shoelace formula. Let me denote coordinates as:

E: (x1, y1) = ((a + 2c)/2, b/2)

N: (x2, y2) = ((a + 4c)/3, b/3)

G: (x3, y3) = ((a + 2c)/3, b/3)

Shoelace formula:

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

Let me compute each term.

First, compute y2 - y3: b/3 - b/3 = 0. So, first term is x1 * 0 = 0.

Second term: x2(y3 - y1) = x2*(b/3 - b/2) = x2*(-b/6)

Third term: x3(y1 - y2) = x3*(b/2 - b/3) = x3*(b/6)

Therefore, Area = | [0 + x2*(-b/6) + x3*(b/6)] / 2 | = | [ (-x2 + x3) * b / 6 ] / 2 | = | [ (-x2 + x3) * b ] / 12 |.

So, compute (-x2 + x3):

x2 = (a + 4c)/3, x3 = (a + 2c)/3. Therefore, -x2 + x3 = - (a + 4c)/3 + (a + 2c)/3 = [ -a -4c + a + 2c ] / 3 = (-2c)/3.

Therefore, Area = | [ (-2c)/3 * b ] / 12 | = | (-2bc)/36 | = | -bc/18 | = bc/18.

So, the area of triangle ENG is bc/18. Now, we need to relate this to the area of triangle ABC, which is given as 144.

Let me compute area of ABC. Coordinates of A(a, b), B(0, 0), C(2c, 0). Using shoelace formula:

Area ABC = | (a*(0 - 0) + 0*(0 - b) + 2c*(b - 0))/2 | = | 0 + 0 + 2c*b / 2 | = | cb | = | bc |. Since area is positive, it's bc. Given that area ABC is 144, so bc = 144.

Therefore, area of triangle ENG is bc/18 = 144 / 18 = 8. So, the area is 8? Wait, that seems straightforward. But let me check my steps again to make sure.

First, coordinates assigned: A(a, b), B(0, 0), C(2c, 0). Then centroid G is ((a + 2c)/3, b/3). Then line through G parallel to BC (which is horizontal) is y = b/3. Found points M and N by intersecting this line with AB and AC. Then found coordinates of E as midpoint of AC. Then applied shoelace formula for triangle ENG. Calculated the area as bc/18, and since area ABC is bc = 144, then bc/18 = 8. So, answer is 8? Wait, but let me check if there's a mistake in coordinates.

Wait, area of ABC: with coordinates A(a, b), B(0, 0), C(2c, 0). The area can also be calculated as base * height / 2. The base BC is 2c units long (from 0 to 2c on x-axis). The height is the y-coordinate of A, which is b. Therefore, area ABC is (2c * b)/2 = bc. Which matches. So bc = 144. Therefore, bc/18 = 144 / 18 = 8. So, area of triangle ENG is 8. Therefore, answer is 8? But let me check with another approach to confirm.

Alternatively, maybe using ratios and properties of centroids and similar triangles.

Since MN is parallel to BC and passes through centroid G. Let me recall that in a triangle, a line parallel to the base divides the sides proportionally. Also, since G is centroid, which divides the medians in ratio 2:1. Let me think about the position of MN.

Since MN is parallel to BC and passes through centroid G. Let me consider the ratio of distances from the base. The centroid is located at 1/3 of the height from the base. Wait, if we consider the height from A to BC, then centroid is located at 1/3 of that height. Therefore, if MN is parallel to BC and passes through centroid, then the distance from MN to BC is 1/3 of the height? Wait, no. Wait, if the centroid is located at 1/3 of the height from the base BC, then the line MN passing through centroid would be located at 1/3 of the height from BC. Therefore, the ratio of lengths of MN to BC would be?

Since the line is parallel to BC, the ratio of lengths is equal to the ratio of distances from A. Wait, if the distance from A to MN is 2/3 of the total height, then MN would be 2/3 of BC? Wait, no. Wait, if the distance from MN to BC is 1/3 of the height, then the distance from A to MN is 2/3 of the height. Therefore, the ratio of similarity is 2/3, so MN = (2/3) BC. Wait, but MN is passing through centroid. Wait, but let me think again.

Let me denote h as the height from A to BC. Then centroid G is located at h/3 from BC. Therefore, the line MN is at height h/3 from BC. Therefore, the distance from A to MN is h - h/3 = 2h/3. Therefore, the ratio of similarity between triangle AMN and triangle ABC is 2/3. Therefore, area of AMN is (2/3)^2 * area ABC = 4/9 * 144 = 64. Therefore, area of AMN is 64. Then, area of ABC is 144, so area of trapezoid MNBC is 144 - 64 = 80. But how does this help with triangle ENG?

Alternatively, maybe I need to find coordinates of E, N, G again. Wait, but earlier calculation gave area of ENG as 8. Let me check with this approach.

Alternatively, since E is midpoint of AC, and N is a point on AC. Let me find the ratio AN / AC. Since MN is parallel to BC, and passes through centroid G. Let me find where N is located on AC.

Earlier, in coordinate terms, N was at ((a + 4c)/3, b/3). Coordinates of A(a, b), C(2c, 0). Let me compute the ratio AN / AC.

Parametrize AC. Let me consider vector from A to C: (2c - a, -b). The coordinates of N are ((a + 4c)/3, b/3). Let me express N as A + t*(vector AC). So:

x-coordinate: a + t*(2c - a) = (a + 4c)/3

Similarly, y-coordinate: b + t*(-b) = b/3

From y-coordinate: b - tb = b/3 => 1 - t = 1/3 => t = 2/3.

Therefore, t = 2/3. Therefore, AN / AC = 2/3. Therefore, N divides AC in ratio AN:NC = 2:1. Wait, but E is the midpoint of AC, so AE:EC = 1:1. Therefore, E is located at 1/2 of AC, and N is located at 2/3 of AC. Therefore, the distance from E to N is (2/3 - 1/2) = 1/6 of AC. Therefore, EN = (1/6) AC.

But how does this help? Maybe we can find the area of ENG. Since G is centroid, and we know coordinates or ratios.

Alternatively, since we know coordinates, maybe we can find vectors or use area ratios.

Alternatively, since we found earlier that area of ENG is bc/18, and bc = 144, so 144/18 = 8. So, answer is 8. But let me check with another approach.

Let me consider mass point or area ratios.

Since MN is parallel to BC and passes through centroid G. As earlier, the ratio of AN / AC = 2/3. Therefore, N divides AC into AN:NC = 2:1. Similarly, M divides AB into AM:MB = 2:1? Wait, earlier in coordinates, M was at (a/3, b/3). Since AB goes from A(a, b) to B(0, 0). So, coordinates of M are (a/3, b/3). Therefore, vector from A to M is (a/3 - a, b/3 - b) = (-2a/3, -2b/3). Therefore, AM is 1/3 of AB? Wait, no. Wait, if you go from A to B, the coordinates of M are (a/3, b/3). So, the ratio AM / AB. Let me parametrize AB. Let me say AB is from A(a, b) to B(0, 0). Let me write parametric equations. Let parameter t go from 0 to 1, with t=0 at A and t=1 at B. Then, coordinates are (a - a*t, b - b*t). So, when does this equal (a/3, b/3)? Let me solve:

a - a*t = a/3 => 1 - t = 1/3 => t = 2/3. Similarly for y-coordinate. Therefore, t = 2/3. Therefore, AM / AB = 2/3. Therefore, AM = 2/3 AB, so MB = 1/3 AB. Therefore, M divides AB in ratio AM:MB = 2:1. Similarly, N divides AC in ratio AN:NC = 2:1.

Therefore, points M and N are located 2/3 along AB and AC respectively from A. Therefore, triangle AMN is similar to triangle ABC with ratio 2/3, so area 4/9 of ABC, which is 4/9 * 144 = 64, as before.

Now, E is the midpoint of AC. So, AE = 1/2 AC. Since AN = 2/3 AC, then EN = AN - AE = 2/3 AC - 1/2 AC = (4/6 - 3/6) AC = 1/6 AC. Therefore, EN = 1/6 AC. So, the length of EN is 1/6 of AC.

But how does this relate to the area of triangle ENG?

Let me think. Since G is centroid, coordinates of G are known. Alternatively, maybe we can find the area of ENG by considering coordinates or ratios.

Alternatively, since we know coordinates of E, N, G, and we found area as 8, which is 144 / 18. Let me check if 18 is related to the ratios. Since area of ABC is 144, and area of ENG is 8, which is 144 / 18. Maybe there's a way to see this ratio.

Alternatively, since EN is 1/6 of AC, and the height from G to EN? Wait, but EN is on AC. Wait, no. Wait, E and N are both on AC, so EN is a segment on AC. Then, triangle ENG has base EN on AC and vertex at G. Therefore, the area of triangle ENG is (1/2)*EN*height, where height is the distance from G to AC.

Wait, but since E and N are on AC, the base is EN, and the height would be the distance from G to line AC. Wait, but if I can compute that distance, then I can compute the area.

Alternatively, since we know coordinates, maybe compute the distance from G to AC.

But let me try this approach. Let me compute the distance from G to AC. Then, area of triangle ENG is 1/2 * EN * distance.

First, find EN. As above, EN = 1/6 AC. The length of AC can be found from coordinates. Coordinates of A(a, b) and C(2c, 0). So, AC has length sqrt[(2c - a)^2 + (0 - b)^2]. But maybe we don't need the actual length, since we can relate areas.

Alternatively, since area of triangle ABC is 144, which is 1/2 * BC * height from A. But maybe we need to relate the distance from G to AC.

Alternatively, since centroid divides the medians in ratio 2:1, the distance from centroid to AC would be 1/3 of the distance from B to AC? Wait, no. Wait, the distance from centroid to AC. Let me think.

The centroid is located 1/3 of the distance from the midpoint of the median. Wait, maybe it's easier to compute using coordinates.

Coordinates of G: ((a + 2c)/3, b/3). Equation of AC: we had earlier y - b = [-b/(2c - a)](x - a). Let me write it in standard form.

Let me rearrange the equation of AC:

y = [-b/(2c - a)](x - a) + b

Multiply both sides by (2c - a):

(2c - a)y = -b(x - a) + b(2c - a)

=> (2c - a)y = -bx + ab + 2bc - ab

=> (2c - a)y = -bx + 2bc

Bring all terms to left:

bx + (2c - a)y - 2bc = 0

So, equation of AC is bx + (2c - a)y - 2bc = 0.

Distance from point G((a + 2c)/3, b/3) to line AC is:

|b*( (a + 2c)/3 ) + (2c - a)*(b/3) - 2bc| / sqrt(b² + (2c - a)²)

Let me compute numerator:

First term: b*(a + 2c)/3

Second term: (2c - a)*(b)/3

Third term: -2bc

So, total numerator:

[ b(a + 2c)/3 + b(2c - a)/3 - 2bc ] = [ b(a + 2c + 2c - a)/3 - 2bc ] = [ b(4c)/3 - 2bc ] = (4bc/3 - 6bc/3) = (-2bc)/3

Absolute value: | -2bc/3 | = 2bc/3

Therefore, distance from G to AC is (2bc/3) / sqrt(b² + (2c - a)^2 )

But the area of triangle ENG is 1/2 * EN * distance.

Earlier, we found EN = 1/6 AC. The length of AC is sqrt[(2c - a)^2 + b²]. Therefore, EN = (1/6) * sqrt[(2c - a)^2 + b²]

Therefore, area of ENG is 1/2 * (1/6) * sqrt(...) * [ (2bc/3) / sqrt(...) ] = 1/2 * 1/6 * 2bc/3 = (1/6)*(bc/3) = bc/18, which matches our previous result. Therefore, area of ENG is bc/18 = 144 / 18 = 8. Therefore, answer is 8.

But let me check once more with another approach. Let me use vectors.

Let me take coordinates again. Let me assume coordinates for simplicity. Let me choose specific coordinates for triangle ABC to make calculations easier. Let me set coordinates such that:

Let me take A(0, 0), B(0, 0)... Wait, no. Let me choose coordinates where calculations are easier. Let me take A(0, 0), B(2, 0), C(0, 2). Wait, but then area of ABC is 2, but we need area 144. Maybe scale later. Let me try.

Let me take A(0, 0), B(2, 0), C(0, 2). Then, area of ABC is 2. Let me compute coordinates of centroid G: ((0 + 2 + 0)/3, (0 + 0 + 2)/3) = (2/3, 2/3). Then, line through G parallel to BC. BC is from (2, 0) to (0, 2), so slope is (2 - 0)/(0 - 2) = -1. Wait, but in this case, BC is not horizontal. Wait, earlier I considered BC horizontal, but here BC has slope -1. Maybe this complicates things. Alternatively, let me take BC horizontal.

Let me take A(0, h), B(-1, 0), C(1, 0). Then, BC is from (-1, 0) to (1, 0), so length 2, and area of ABC is (base * height)/2 = (2 * h)/2 = h. Let me set h = 144, but maybe take h = 1 for simplicity, then scale later. Wait, but maybe this is getting too convoluted.

Alternatively, take A(0, 3), B(0, 0), C(2, 0). Then, area of ABC is (base * height)/2 = (2 * 3)/2 = 3. Then centroid G is ((0 + 0 + 2)/3, (3 + 0 + 0)/3) = (2/3, 1). Then, line through G parallel to BC. Since BC is horizontal (from (0,0) to (2, 0)), so line through G is horizontal line y = 1. This intersects AB and AC.

Equation of AB: from (0, 3) to (0, 0). Wait, AB is vertical line x = 0. Wait, but line y = 1 intersects AB at (0, 1). Similarly, equation of AC: from (0, 3) to (2, 0). Slope is (0 - 3)/(2 - 0) = -3/2. Equation: y - 3 = -3/2 (x - 0) => y = -3/2 x + 3. Intersection with y = 1:

1 = -3/2 x + 3 => -3/2 x = -2 => x = (-2) * (-2/3) = 4/3. Therefore, point N is (4/3, 1). So, points M(0, 1) and N(4/3, 1). E is midpoint of AC: coordinates of A(0, 3) and C(2, 0), so midpoint E is ((0 + 2)/2, (3 + 0)/2) = (1, 1.5). G is (2/3, 1). So, triangle ENG has coordinates E(1, 1.5), N(4/3, 1), G(2/3, 1). Let me compute its area.

Using shoelace formula:

Coordinates:

E(1, 1.5), N(4/3, 1), G(2/3, 1)

Compute area:

Arrange points:

E(1, 3/2), N(4/3, 1), G(2/3, 1)

Compute using formula:

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

Plugging in:

x1 = 1, y1 = 3/2

x2 = 4/3, y2 = 1

x3 = 2/3, y3 = 1

Compute:

1*(1 - 1) + (4/3)*(1 - 3/2) + (2/3)*(3/2 - 1)

= 1*(0) + (4/3)*(-1/2) + (2/3)*(1/2)

= 0 - 4/6 + 2/6 = (-4 + 2)/6 = -2/6 = -1/3

Take absolute value and divide by 2? Wait, no. Wait, the formula is |sum / 2|. Wait, the formula is |sum| / 2. Wait, the formula is:

Area = | [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] / 2 |.

So, the sum inside is -1/3, so area is | -1/3 / 2 |? Wait no, the entire expression is divided by 2. Wait, no:

Wait, the formula is:

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

So, the sum inside is -1/3, so area is | (-1/3) / 2 |? Wait, no. Wait, the entire expression is divided by 2. Wait, no, the formula is:

It's [sum] divided by 2. So, the sum is -1/3, so area is | -1/3 / 2 |? Wait, no, no. Wait, the formula is:

Area = | [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] / 2 |.

So, the numerator is -1/3, so area is | (-1/3) / 2 |? No, no. Wait, the numerator is already calculated as -1/3. Wait, no:

Wait, let me recalculate:

Original calculation:

x1(y2 - y3) = 1*(1 - 1) = 0

x2(y3 - y1) = (4/3)*(1 - 3/2) = (4/3)*(-1/2) = -2/3

x3(y1 - y2) = (2/3)*(3/2 - 1) = (2/3)*(1/2) = 1/3

Therefore, total sum is 0 - 2/3 + 1/3 = -1/3. Then divide by 2: (-1/3)/2 = -1/6. Take absolute value: 1/6. Therefore, area is 1/6.

But in this case, area of ABC is 3, and area of ENG is 1/6. Therefore, ratio is 1/6 divided by 3 = 1/18. Therefore, area of ENG is (1/18)*area ABC. Therefore, if area ABC is 144, then area ENG is 144 / 18 = 8. Which matches our previous result. Therefore, this confirms that the area is indeed 8. Therefore, the answer is 8.

Therefore, I think the answer is \boxed{8}.

**Final Answer**
\boxed{8}

To solve the problem, we are given a triangle $ ABC $ with medians $ AD $, $ BE $, and $ CF $ intersecting at the centroid $ G $. A line through $ G $, parallel to $ BC $, intersects $ AB $ at $ M $ and $ AC $ at $ N $. We are to find the area of triangle $ ENG $, where $ E $ is the midpoint of $ AC $, and the area of triangle $ ABC $ is given as 144.

---

### Step 1: Coordinate Setup

To simplify the geometry, we place triangle $ ABC $ in the coordinate plane:

- Let $ A = (0, 0) $
- Let $ B = (2b, 0) $
- Let $ C = (2c, 2d) $

This setup ensures that the base $ BC $ lies along the x-axis, and the triangle is symmetric in a way that simplifies calculations.

---

### Step 2: Centroid Coordinates

The centroid $ G $ of triangle $ ABC $ is the average of the coordinates of the vertices:

$$
G = \left( \frac{0 + 2b + 2c}{3}, \frac{0 + 0 + 2d}{3} \right) = \left( \frac{2b + 2c}{3}, \frac{2d}{3} \right)
$$

---

### Step 3: Line Through $ G $ Parallel to $ BC $

Since $ BC $ lies along the x-axis, the line through $ G $ that is parallel to $ BC $ is horizontal. Therefore, its equation is:

$$
y = \frac{2d}{3}
$$

This line intersects:

- **Line $ AB $** at point $ M $
- **Line $ AC $** at point $ N $

---

### Step 4: Coordinates of Points $ M $, $ N $, and $ E $

#### Point $ M $ on $ AB $:

Line $ AB $ goes from $ A = (0, 0) $ to $ B = (2b, 0) $, so it's horizontal. The line $ y = \frac{2d}{3} $ intersects $ AB $ at:

$$
M = \left( \frac{2b}{3}, \frac{2d}{3} \right)
$$

#### Point $ N $ on $ AC $:

Line $ AC $ goes from $ A = (0, 0) $ to $ C = (2c, 2d) $. The slope of $ AC $ is:

$$
m = \frac{2d - 0}{2c - 0} = \frac{d}{c}
$$

So the equation of line $ AC $ is:

$$
y = \frac{d}{c}x
$$

Setting $ y = \frac{2d}{3} $, we solve for $ x $:

$$
\frac{2d}{3} = \frac{d}{c}x \Rightarrow x = \frac{2c}{3}
$$

So, point $ N = \left( \frac{2c}{3}, \frac{2d}{3} \right) $

#### Point $ E $, the midpoint of $ AC $:

$$
E = \left( \frac{0 + 2c}{2}, \frac{0 + 2d}{2} \right) = \left( c, d \right)
$$

---

### Step 5: Area of Triangle $ ENG $

We now have the coordinates of the three points:

- $ E = (c, d) $
- $ N = \left( \frac{2c}{3}, \frac{2d}{3} \right) $
- $ G = \left( \frac{2b + 2c}{3}, \frac{2d}{3} \right) $

We apply the **shoelace formula** for the area of triangle $ ENG $:

$$
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
$$

Substitute the coordinates:

- $ x_1 = c, y_1 = d $
- $ x_2 = \frac{2c}{3}, y_2 = \frac{2d}{3} $
- $ x_3 = \frac{2b + 2c}{3}, y_3 = \frac{2d}{3} $

Compute each term:

- $ x_1(y_2 - y_3) = c \left( \frac{2d}{3} - \frac{2d}{3} \right) = 0 $
- $ x_2(y_3 - y_1) = \frac{2c}{3} \left( \frac{2d}{3} - d \right) = \frac{2c}{3} \left( -\frac{d}{3} \right) = -\frac{2cd}{9} $
- $ x_3(y_1 - y_2) = \frac{2b + 2c}{3} \left( d - \frac{2d}{3} \right) = \frac{2b + 2c}{3} \cdot \frac{d}{3} = \frac{(2b + 2c)d}{9} $

Add them:

$$
\text{Area} = \frac{1}{2} \left| 0 - \frac{2cd}{9} + \frac{(2b + 2c)d}{9} \right| = \frac{1}{2} \left| \frac{2bd + 2cd - 2cd}{9} \right| = \frac{1}{2} \cdot \frac{2bd}{9} = \frac{bd}{9}
$$

---

### Step 6: Relating to Area of Triangle $ ABC $

The area of triangle $ ABC $ is:

$$
\text{Area}_{ABC} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2b \cdot 2d = 2bd
$$

Given that $ \text{Area}_{ABC} = 144 $, we have:

$$
2bd = 144 \Rightarrow bd = 72
$$

Now, the area of triangle $ ENG $ is:

$$
\text{Area}_{ENG} = \frac{bd}{9} = \frac{72}{9} = 8
$$

---

### Final Answer

$$
\boxed{8}
$$

Solution:
Since $E$ is the midpoint of $AC$, the area of triangle $BCE$ is half the area of triangle $ABC$, or $144/2 = 72$.

[asy]
import geometry;

unitsize(1 cm);

pair A, B, C, D, E, F, G, M, N;

A = (1,3);
B = (0,0);
C = (4,0);
D = (B + C)/2;
E = (C + A)/2;
F = (A + B)/2;
G = (A + B + C)/3;
M = extension(G, G + B - C, A, B);
N = extension(G, G + B - C, A, C);

draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);
draw(M--N);

label("$A$", A, dir(90));
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$G$", G, SSW);
label("$M$", M, NW);
label("$N$", N, NE);
[/asy]

Since $GN$ is parallel to $BC$, triangles $ENG$ and $ECB$ are similar.  Furthermore, $G$ is the centroid of triangle $ABC$, so the ratio of similarity is $EG/EB = 1/3$.  Therefore, the area of triangle $ENG$ is $72 \cdot (1/3)^2 = \boxed{8}$.