Question

$ABCD$ is a parallelogram. $L$ is a point on $BC$ which divides $BC$ in the ratio $1:2$. $AL$ intersects $BD$ at $P$. $M$ is a point on $DC$ which divides $DC$ in the ratio $1:2$ and AM intersects $BD$ in $Q$.

Point $P$ divides $AL$ in the ratio

• A. $1:2$
• B. $1:3$
• C. $3:1$
• D. $2:1$

Answer 1

Answer: (C)

$3:1$

Given, $BL:LC=1:2$ and $DM:MC=1:2$

Now, $BL:BC=1:3$
Consider $△APD$ and $△LPB$

$\angle ADP=\angle LBP$

$\angle APD=\angle BPL$

By AA corollary,

$△APD\sim △LPB$

$\Rightarrow$ Corresponding parts are proportional.

$\Rightarrow \frac{AP}{LP}=\frac{AD}{BL}$
$\Rightarrow \frac{AP}{LP}=\frac{BC}{BL}$ ($\because AD=BC$)
$\Rightarrow \frac{AP}{LP}=3:1$