Question

The median $AD$ of a triangle $ABC$ is bisected at $E,BE$ meets $AC$ in $F$; the $AF:AC=$

• A. $\frac{3}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{3}$

Let $\frac{AF}{FC}=\frac{\lambda }{1}$

and let $D(0,0),B(-a,0),C(a,0),A(h,k)$, then

$E=(\frac{h}{2},\frac{k}{2})$ and

$F=(\frac{\lambda a+h}{\lambda +1},\frac{k}{\lambda +1})$

Now, $B,E,F$ are collonear

$\Rightarrow \left|\begin{array}{ccc}\frac{\lambda a+h}{\lambda +1}& \frac{k}{\lambda +1}& 1\\ \frac{h}{2}& \frac{k}{2}& 1\\ -a& 0& 1\end{array}\right|=0\Rightarrow \lambda =\frac{1}{2}\Rightarrow \frac{AF}{AC}=\frac{1}{3}$