Question

$P$ is the midpoint of side $BC$ of $△ABC$, $Q$ is the midpoint of $AP$, $BQ$ when produced meets $AC$ at $L$. prove $AL=\frac{1}{3}AC$.

Answer 1

In $△BCL$, we have $PM\parallel BL$

$\therefore$ by using Basic Proportionality theorem, we have

$\frac{CP}{PB}=\frac{CM}{ML}$ .......$\left(1\right)$

But $BP=PC$

$\Rightarrow \frac{PC}{BP}=1$ ........$\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$ we get

$1=\frac{CM}{ML}$

$\Rightarrow CM=ML$ ...........$\left(3\right)$

Now, in $△APM,$ we have

$QL\parallel PM$

$\therefore$, by Basic Proportionality theorem, we have

$\frac{AQ}{QP}=\frac{AL}{LM}$ ......$\left(4\right)$

But $AQ=QP$

$\Rightarrow \frac{AQ}{QP}=1$ ......$\left(5\right)$

Comparing $\left(4\right)$ and $\left(5\right)$ we get

$1=\frac{AL}{LM}$

$\Rightarrow AL=LM$ ......$\left(6\right)$

Comparing $\left(3\right)$ and $\left(6\right)$, we have

$CM=ML=AL$

Now,$AC=AL+LM+MC=3AL$

$\Rightarrow AL=\frac{1}{3}AC$