The median $AD$ of a triangle $ABC$ is bisected at $E$ and $BE$ meets $AC$ at $F$; then $AF:AC =$
Step-by-Step Solution
Key Concept: Collinearity of three points is determined using the determinant condition, which constrains the parameter $\lambda$.
With $D(0,0)$, $B(-a,0)$, $C(a,0)$, $A(h,k)$, and $\frac{AF}{FC} = \lambda$, we find $E = \left(\frac{h}{2}, \frac{k}{2}\right)$ and $F = \left(\frac{\lambda a + h}{\lambda + 1}, \frac{k}{\lambda+1}\right)$. For collinearity of $B$, $E$, $F$, the determinant condition yields $\lambda = \frac{1}{2}$, giving $\frac{AF}{AC} = \frac{1}{3}$.
Correct Answer: 2