The median AD of the triangle ABC is bisected at E and BE meets AC at F. Find AF:FC.

Options

A. `3//4`
B. `1//3`
C. `1//2`
D. `1//4`

Explanation

To solve the problem of finding the ratio \( AF:FC \) in triangle \( ABC \) where \( AD \) is the median bisected at \( E \) and \( BE \) meets \( AC \) at \( F \), we can follow these steps:

### Step 1: Set up the coordinates
Let’s assign coordinates to the points:
- Let \( B = (-a, 0) \)
- Let \( C = (a, 0) \)
- Let \( A = (h, k) \)

### Step 2: Find the coordinates of point \( D \)
Point \( D \) is the midpoint of \( BC \):
\[ D = \left( \frac{-a + a}{2}, \frac{0 + 0}{2} \right) = (0, 0) \]

### Step 3: Find the coordinates of point \( E \)
Point \( E \) is the midpoint of \( AD \):
\[ E = \left( \frac{h + 0}{2}, \frac{k + 0}{2} \right) = \left( \frac{h}{2}, \frac{k}{2} \right) \]

### Step 4: Express the coordinates of point \( F \)
Let \( AF:FC = \lambda:1 \). Thus, we can express the coordinates of point \( F \) on line segment \( AC \):
\[ F = \left( \frac{\lambda a + h}{\lambda + 1}, \frac{\lambda \cdot 0 + k}{\lambda + 1} \right) = \left( \frac{\lambda a + h}{\lambda + 1}, \frac{k}{\lambda + 1} \right) \]

### Step 5: Check collinearity of points \( B, E, F \)
For points \( B, E, F \) to be collinear, the area of triangle \( BEF \) must be zero. We can use the determinant method:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| = 0 \]
Substituting the coordinates:
\[ \frac{1}{2} \left| (-a)\left(\frac{k}{\lambda + 1} - \frac{k}{2}\right) + \frac{h}{2}\left(0 - 0\right) + \frac{\lambda a + h}{\lambda + 1}\left(0 - \frac{k}{2}\right) \right| = 0 \]
This simplifies to:
\[ -a \left(\frac{2k - k(\lambda + 1)}{2(\lambda + 1)}\right) + \left(\frac{\lambda a + h}{\lambda + 1}\right)\left(-\frac{k}{2}\right) = 0 \]

### Step 6: Solve for \( \lambda \)
By simplifying the above equation, we can find the value of \( \lambda \). After solving, we find:
\[ \lambda = \frac{1}{2} \]

### Step 7: Find the ratio \( AF:FC \)
Since \( AF:FC = \lambda:1 \):
\[ AF:FC = \frac{1}{2}:1 = 1:2 \]

### Final Answer
Thus, the ratio \( AF:FC \) is \( 1:2 \).

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