<p><strong>Example 30.</strong> The median AD of the △ABC is bisected at E. BE meets AC in F. Then, AF : AC is equal to</p>
Step-by-Step Solution
Key Concept: Use position vectors to express points in terms of reference points. Apply collinearity conditions to find the ratio in which F divides AC.
Solution: Let position vector of A w.r.t. B is \(\vec{a}\) and that of C w.r.t. B is \(\vec{c}\). Position vector of D w.r.t. B = \(\frac{0 + \vec{c}}{2} = \frac{\vec{c}}{2}\) Position vector of E = \(\frac{\vec{a} + \frac{\vec{c}}{2}}{2} = \frac{\vec{a}}{2} + \frac{\vec{c}}{4}\) Let AF : FC = \(\lambda : 1\) and position vector of F = \(\frac{\lambda \vec{c} + \vec{a}}{1 + \lambda}\) Since BE meets AC in F, using the condition that B, E, F are collinear and solving the system of equations, we get \(\lambda = \frac{1}{2}\). Therefore, AF : AC = 1 : 3, so the answer is (b) 1/3.
Correct Answer: B