Consider a square ABCD of side length 16. Let E, F be points on CD such that CE = EF = FD. Let the line BF and AE meet in M. The area of triangle MAB is

To solve the problem, let's first visualize the square \(ABCD\) with side length \(16\), and place it on a coordinate plane:

- Assume point \(A = (0, 16)\), \(B = (16, 16)\), \(C = (16, 0)\), and \(D = (0, 0)\).

Points \(E\) and \(F\) divide side \(CD\) into three equal segments:

1. Since \(CD = 16\), each segment \(CE = EF = FD = \frac{16}{3} \approx 5.33\).
   
2. The coordinates for \(E\) and \(F\) are:
   - \(E = \left(16 - \frac{16}{3}, 0\right) = \left(\frac{32}{3}, 0\right)\)
   - \(F = \left(16 - 2 \times \frac{16}{3}, 0\right) = \left(\frac{16}{3}, 0\right)\)

### Step 1: Find Equations of Lines AE and BF

**Line \(AE\):**

1. Point \(A = (0, 16)\) and \(E = \left(\frac{32}{3}, 0\right)\).
2. Slope of line \(AE\) is:
   \[
   \text{slope of } AE = \frac{0 - 16}{\frac{32}{3} - 0} = -\frac{16}{\frac{32}{3}} = -\frac{16 \cdot 3}{32} = -\frac{3}{2}
   \]
3. The equation of line \(AE\) in point-slope form is:
   \[
   y - 16 = -\frac{3}{2}x \implies y = -\frac{3}{2}x + 16
   \]

**Line \(BF\):**

1. Point \(B = (16, 16)\) and \(F = \left(\frac{16}{3}, 0\right)\).
2. Slope of line \(BF\) is:
   \[
   \text{slope of } BF = \frac{0 - 16}{\frac{16}{3} - 16} = \frac{-16}{\frac{16}{3} - 16} = \frac{-16}{\frac{16 - 48}{3}} = \frac{-16 \cdot 3}{-32} = \frac{48}{32} = \frac{3}{2}
   \]
3. The equation of line \(BF\) in point-slope form is:
   \[
   y - 16 = \frac{3}{2}(x - 16) \implies y = \frac{3}{2}x - 8
   \]

### Step 2: Find Coordinates of Point M

To find the intersection \(M\) of lines \(AE\) and \(BF\), set their equations equal:

\[
-\frac{3}{2}x + 16 = \frac{3}{2}x - 8
\]

Combine like terms:

\[
16 + 8 = \frac{3}{2}x + \frac{3}{2}x \implies 24 = 3x \implies x = 8
\]

Substitute \(x = 8\) back into either line equation to find \(y\):

\[
y = -\frac{3}{2} \cdot 8 + 16 = -12 + 16 = 4
\]

Thus, point \(M = (8, 4)\).

### Step 3: Calculate the Area of Triangle MAB

The vertices of triangle \(MAB\) are \(M = (8, 4)\), \(A = (0, 16)\), and \(B = (16, 16)\).

The area of a triangle given by coordinates \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is:

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

Substitute the coordinates:

\[
\text{Area} = \frac{1}{2} \left| 8(16 - 16) + 0(16 - 4) + 16(4 - 16) \right|
\]

Simplify:

\[
\text{Area} = \frac{1}{2} \left| 0 + 0 + 16(-12) \right| = \frac{1}{2} \left| -192 \right| = \frac{192}{2} = 96
\]

### Conclusion

The area of triangle \(MAB\) is \(\boxed{96}\).

Learn how to find the area of triangle MAB in a square ABCD with side length 16, where points E and F divide side CD into equal segments. This problem involves coordinate geometry, algebra, and the triangle area formula, suitable for Grades 10-12.

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