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 $\boxed{96}$.