Question

If the median AD of the triangle ABC is bisected at E and BE meets AC in F, then AF : AC =

• A. \(1 : 4 \)
• B. \(1 : 3 \)
• C. \(1 : 2 \)
• D. \(3 : 4 \)

Hint:

Problems involving medians and intersecting line segments in a triangle often use Ceva's Theorem or Menelaus' Theorem. Alternatively, constructing a parallel line through a known midpoint can simplify the problem by creating similar triangles or allowing the application of the midpoint theorem, breaking down complex ratios into simpler ones.

Answer 1

The Correct Option is B

Step 1: Draw the triangle and label the given points.
Let ABC be a triangle.
AD is a median, so D is the midpoint of BC (\(BD = DC\)).
E is the midpoint of AD (\(AE = ED\)).
The line segment BE is extended to meet AC at point F. 
Step 2: Apply Menelaus' Theorem.
Consider triangle ADC and the transversal line segment B-E-F. The points F, E, B are collinear. F is on AC, E is on AD. B is on the line containing CD (or BC).
According to Menelaus' Theorem for triangle ADC and transversal B-E-F: \[ \left(\frac{AF}{FC}\right) \times \left(\frac{CB}{BD}\right) \times \left(\frac{DE}{EA}\right) = 1 \] From the given information:
D is the midpoint of BC, so \(BC = BD + DC\). Since \(BD = DC\), we have \(BC = 2BD\).
Therefore, \(\frac{CB}{BD} = \frac{2BD}{BD} = 2\). E is the midpoint of AD, so \(AE = ED\). Therefore, \(\frac{DE}{EA} = \frac{ED}{ED} = 1\). Substitute these values into the Menelaus' Theorem equation: \[ \left(\frac{AF}{FC}\right) \times (2) \times (1) = 1 \] \[ 2 \times \frac{AF}{FC} = 1 \] \[ \frac{AF}{FC} = \frac{1}{2} \] 
Step 3: Determine the ratio AF : AC.
From \(\frac{AF}{FC} = \frac{1}{2}\), we have \(FC = 2 \times AF\). We want the ratio \(AF : AC\). We know that \(AC = AF + FC\). Substitute \(FC = 2AF\) into the equation for AC: \[ AC = AF + 2AF \] \[ AC = 3AF \] Therefore, the ratio \(AF : AC\) is: \[ \frac{AF}{AC} = \frac{AF}{3AF} = \frac{1}{3} \] So, \(AF : AC = 1 : 3\). 
Alternative Method (using parallel lines and midpoint theorem):
1. Draw a line through D parallel to BF, let it intersect AC at G. So \(DG \parallel BF\).
2. In \(\triangle BCF\), D is the midpoint of BC (\(BD=DC\)). Since \(DG \parallel BF\), by the converse of midpoint theorem (or Thales's theorem), G must be the midpoint of CF.
So, \(CG = GF\).
3. In \(\triangle ADG\), E is the midpoint of AD (\(AE=ED\)). 
Since \(EF\) is a segment of BE, and \(DG \parallel BE\), it implies \(EF \parallel DG\). 
By the converse of midpoint theorem, F must be the midpoint of AG. So, \(AF = FG\).
4. Combining the results: \(AF = FG\) and \(FG = GC\).
Therefore, \(AF = FG = GC\).
5. Now, \(AC = AF + FG + GC = AF + AF + AF = 3AF\).
So, \(AF : AC = AF : 3AF = 1 : 3\).
The final answer is $\boxed{1 : 3}}$.