Question

In the following figure, $M$ is mid-point of $BC$ of a parallelogram $ABCD$. $DM$ intersects the diagonal $AC$ at $P$ and $AB$ is produced at $E$. Then

• A. $PE=3PD$
• B. $PE=4PD$
• C. $PE=2PD$
• D. $PE=PD$

Answer 1

The correct option is C $PE=2PD$

Given:

$ABCD$ is a parallelogram.

$M$ is mid point of $BC$.
In $△DMC$ and $△EMB,$
$\angle DMC=\angle BME$ $($Vertically opposite angles$)$
$\angle DCM=\angle MBE$ $($Alternate angles$)$
$MC=MB$ $(M$ is mid point of $BC)$

Thus, $△DMC\cong △EMB$ $(ASA$ rule$)$

Thus, $DC=BE$ $($Since corresponding parts of congruent triangles are congruent.$)$
Now, In $△DPC$ and $△APE,$

$\angle DPC=\angle APE$ $($Vertically opposite angles$)$
$\angle CDP=\angle AEP$ $($Alternate angles$)$
$\angle PCD=\angle PAE$ $($Alternate angles$)$
Thus, $△DPC\sim △EPA$ $(AAA$ rule$)$

Hence, $\frac{PE}{PD}=\frac{AE}{CD}$

$\Rightarrow \frac{PE}{PD}=\frac{AB+BE}{CD}$ (Since $AB=BE=CD$)

$\Rightarrow \frac{PE}{PD}=2$

$\Rightarrow PE=2PD$

Hence, option C is correct.