In a $ riangle ABC$, AD is the median through A and E is the mid point of AD, and BE produced meets AC at F, then AF is equal to :

Q: In a $\triangle ABC$, AD is the median through A and E is the mid point of AD, and BE produced meets AC at F, then AF is equal to :

In a $\triangle ABC$, AD is the median through A and E is the mid point of AD, and BE produced meets AC at F, then AF is equal to :

Question

In a $	riangle ABC$, AD is the median through A and E is the mid point of AD, and BE produced meets AC at F, then AF is equal to :

Accepted Answer

To solve this problem, we need to understand the geometric configuration and apply some properties of medians and midpoints in a triangle.

### Explanation for Option 3 being correct:

1. **Understanding the Configuration:**
   - In triangle $ \triangle ABC $, AD is a median, meaning D is the midpoint of BC.
   - E is the midpoint of AD, which implies that AE = ED.
   - BE is extended to meet AC at F.

2. **Using the Properties of Medians and Midpoints:**
   - Since E is the midpoint of AD, AE = ED = $\frac{1}{2}$AD.
   - In a triangle, if a line is drawn from a vertex to the midpoint of the opposite side, it divides the triangle into two smaller triangles of equal area.

3. **Applying Menelaus' Theorem:**
   - Menelaus' theorem can be applied to triangle $ \triangle ACD $ with transversal BEF.
   - According to Menelaus' theorem, for points B, E, and F to be collinear, the following relation must hold:
     $$
     \frac{AB}{BC} \cdot \frac{CD}{DA} \cdot \frac{AF}{FC} = 1
     $$
   - Since D is the midpoint of BC, $ \frac{CD}{DA} = 1 $.
   - E is the midpoint of AD, so $ \frac{AE}{ED} = 1 $.

4. **Solving for AF:**
   - By substituting the known ratios into Menelaus' theorem, we simplify to find:
     $$
     \frac{AF}{FC} = \frac{1}{2}
     $$
   - This implies that AF is one-third of AC because:
     $$
     AF = \frac{1}{3}AC
     $$

Thus, the correct answer is Option 3: $ \frac{AC}{3} $.

### Explanation for why other options are incorrect:

1. **Option 1: $\frac{AC}{4}$**
   - This option suggests that AF is one-fourth of AC, which does not align with the geometric properties and the application of Menelaus' theorem in this configuration.

2. **Option 2: $\frac{AC}{4}$**
   - This is identical to Option 1 and is incorrect for the same reasons.

3. **Option 4: $\frac{AC}{2}$**
   - This option implies that AF is half of AC, which contradicts the result obtained from Menelaus' theorem and the properties of medians and midpoints.

In conclusion, the correct answer is Option 3: $ \frac{AC}{3} $, as justified by the application of geometric principles and Menelaus' theorem.