The median AD of ΔABC is bisected at E and BE is extended to meet the side AC in F. The ratio AF : FC = ?
Use mass points (or Menelaus' theorem). D is the midpoint of BC, and E is the midpoint of AD.
Menelaus on triangle ADC with transversal B-E-F: \(\frac{AE}{ED} \cdot \frac{DB}{BC} \cdot \frac{CF}{FA} = 1\)
\(1 \cdot \frac{1}{2} \cdot \frac{CF}{FA} = 1 \Rightarrow \frac{CF}{FA} = 2\)
AF : FC = 1 : 2
Answer: (C)
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