Using vectors show that $AL$ bisects $BC$. A line $EF$ drawn parallel to the base $BC$ of a 
$∆ ABC$ meets $AB$ & $AC$ in $F$ & $E$ respectively. $BE$ & $CF$
meet in $L$. Using vectors we have to show that $AL$ bisects $BC$.

I tried as 

Let $B$ is origin and $A$ is vector $\vec a$ and $C$ is vector $\vec c$.
Now let $\frac{AF}{AB} = m$.
Then vectors $\vec f$ and $\vec e$ are respectively $(1-m)\vec a$ and $(1+m)\vec a-m\vec c$.
But now how to proceed . 


Then vectors $\vec f$ and $\vec e$ are respectively $(1-m)\vec a$ and $(1+m)\vec a-m\vec c$.

It should be $\vec e=(1\color{red}{-}m)\vec a\color{red}{+}m\vec c$.

There exist real numbers $s,t$ such that 
$$\vec{l}=s\vec e=(1-m)s\vec a+ms\vec c\quad\text{and}\quad \vec l=t\vec f+(1-t)\vec c=(1-m)t\vec a+(1-t)\vec c$$
to have
$$(1-m)s=(1-m)t\quad\text{and}\quad ms=1-t$$
so
$$s=t=\frac{1}{m+1}$$
Therefore, we get
$$\vec l=\frac{1-m}{m+1}\vec a+\frac{m}{m+1}\vec c$$
which can be written as
$$\vec{AL}=\frac{m}{m+1}(\vec{AB}+\vec{AC})$$
The claim follows from this.
