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

We are tasked with finding the area of triangle \( MAB \), where:

- \( ABCD \) is a square with side length \( 16 \),
- \( E \) and \( F \) are points on side \( CD \) such that \( CE = EF = FD \), i.e., the points divide \( CD \) into three equal segments,
- \( BF \) and \( AE \) are lines that intersect at point \( M \).

### Step 1: Coordinates of key points
- Let the square \( ABCD \) be on the coordinate plane with \( A = (0, 16) \), \( B = (16, 16) \), \( C = (16, 0) \), and \( D = (0, 0) \).
- Since \( E \) and \( F \) divide \( CD \) into three equal parts, we have:
  - \( E = \left( \frac{2}{3} \times 16, 0 \right) = (10.67, 0) \),
  - \( F = \left( \frac{1}{3} \times 16, 0 \right) = (5.33, 0) \).

### Step 2: Equations of lines \( AE \) and \( BF \)
- The slope of line \( AE \) (connecting \( A = (0, 16) \) and \( E = (10.67, 0) \)) is:
  \[
  \text{slope of } AE = \frac{0 - 16}{10.67 - 0} = -\frac{16}{10.67} \approx -1.5
  \]
  So, the equation of line \( AE \) is:
  \[
  y - 16 = -1.5(x - 0) \quad \Rightarrow \quad y = -1.5x + 16
  \]
- The slope of line \( BF \) (connecting \( B = (16, 16) \) and \( F = (5.33, 0) \)) is:
  \[
  \text{slope of } BF = \frac{0 - 16}{5.33 - 16} = \frac{-16}{-10.67} \approx 1.5
  \]
  So, the equation of line \( BF \) is:
  \[
  y - 16 = 1.5(x - 16) \quad \Rightarrow \quad y = 1.5x - 8
  \]

### Step 3: Finding the intersection point \( M \)
To find the coordinates of point \( M \), we set the equations of \( AE \) and \( BF \) equal:
\[
-1.5x + 16 = 1.5x - 8
\]
Solving for \( x \):
\[
16 + 8 = 1.5x + 1.5x \quad \Rightarrow \quad 24 = 3x \quad \Rightarrow \quad x = 8
\]
Substitute \( x = 8 \) into the equation of \( AE \) to find \( y \):
\[
y = -1.5(8) + 16 = -12 + 16 = 4
\]
So, \( M = (8, 4) \).

### Step 4: Area of triangle \( MAB \)
The area of triangle \( MAB \) can be found using the formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates \( A(0, 16) \), \( B(16, 16) \), and \( M(8, 4) \):
\[
\text{Area} = \frac{1}{2} \left| 0(16 - 4) + 16(4 - 16) + 8(16 - 16) \right|
\]
\[
= \frac{1}{2} \left| 0 + 16(-12) + 8(0) \right| = \frac{1}{2} \left| -192 \right| = \frac{192}{2} = 96
\]

Thus, the area of triangle \( MAB \) is \( \boxed{96} \).

Do you have any questions or want details on a specific part of this solution? Here are 5 related questions:
1. How does the division of \( CD \) into equal parts affect the result?
2. Can we find the area of triangle \( MAB \) using a different method?
3. What is the geometric significance of the intersection of \( AE \) and \( BF \)?
4. How would the result change if the square had a different side length?
5. How can the coordinates of intersection points be applied to other geometric problems?

**Tip:** When finding the area of a triangle from its vertex coordinates, always verify the orientation of the points to avoid sign errors in the formula.

Discover how to calculate the area of triangle MAB formed by the intersection of lines AE and BF inside a square of side length 16. The solution explores the use of slope formulas and the area formula for triangles in coordinate geometry. Suitable for students in Grades 9-12.

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