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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 .

1 Answers 1

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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.

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    Why e should be like that – 2017-01-21
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    @koolman: Because $\vec{e}=\vec a+\vec{AE}=\vec a+m\vec{AC}=\vec a+m(\vec c-\vec a)=(1-m)\vec a+m\vec c$. Another way is $\vec e=\vec f+\vec{FE}=(1-m)\vec a+m\vec c$. – 2017-01-21
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    What is vector l and how you hvae written its second equation – 2017-01-21
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    @koolman: $\vec l=\vec{BL}$. So, the line means that $\vec{BL}=s\vec{BE}$ and $\vec{BL}=t\vec{BF}+(1-t)\vec{BC}$. – 2017-01-21
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    @koolman: $L$ is on the line $BE$, so there exists a real number $s$ such that $\vec{BL}=s\vec{BE}$. Also, $L$ is on the line $CF$, so there exists a real number $t$ such that $\vec{BL}=t\vec{BF}+(1-t)\vec{BC}$. – 2017-01-21
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    From the result how can we say AL bisects BC – 2017-01-21
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    @koolman: Let $G$ be a point such that $\vec{AB}+\vec{AC}=\vec{AG}$. Then, we know that $ABGC$ is a parallelogram. Let $M$ be the midpoint of the line segment $BC$. Then, we know that $M$ is on the line $AG$. – 2017-01-21
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    Thanks a lot , nice explanation – 2017-01-21
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    @koolman: You are welcome. – 2017-01-21