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:

Place the square $$ABCD$$ on the coordinate plane with
$$A(0,0),\;B(16,0),\;C(16,16),\;D(0,16).$$

The side $$CD$$ is the segment from $$(16,16)$$ to $$(0,16).$$
Since $$CE=EF=FD=\dfrac{16}{3},$$ the trisection points are

$$E\left(16-\dfrac{16}{3},\,16\right)=\left(\dfrac{32}{3},\,16\right),\qquad F\left(16-\dfrac{32}{3},\,16\right)=\left(\dfrac{16}{3},\,16\right).$$

Equation of $$BF$$
Slope $$m_{BF}=\dfrac{16-0}{\tfrac{16}{3}-16} =\dfrac{16}{-\tfrac{32}{3}}=-\dfrac{3}{2}.$$ Hence $$BF: y=-\dfrac{3}{2}(x-16).$$

Equation of $$AE$$
Slope $$m_{AE}=\dfrac{16-0}{\tfrac{32}{3}-0} =\dfrac{16}{\tfrac{32}{3}}=\dfrac{3}{2}.$$ Thus $$AE: y=\dfrac{3}{2}x.$$

Intersection point $$M = BF \cap AE$$
Set the two expressions for $$y$$ equal: $$\dfrac{3}{2}x=-\dfrac{3}{2}x+24 \;\Longrightarrow\; 3x=24 \;\Longrightarrow\; x=8.$$ Then $$y=\dfrac{3}{2}\times8=12.$$ So $$M(8,12).$$

Area of $$\triangle MAB$$
Base $$AB$$ lies on the $$x$$-axis from $$x=0$$ to $$x=16,$$ so $$AB=16.$$ The perpendicular distance of $$M$$ from $$AB$$ is its $$y$$-coordinate $$12.$$ Therefore

$$ \text{Area}=\dfrac12\,(\text{base})\,(\text{height}) =\dfrac12\,(16)\,(12)=96. $$

Hence the required area is 96.