Question

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

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

Answer 1

The correct option is B $PE=2PD$

Given: ABCD is a parallelogram. M is mid point of BC.
In $△DMC$ and $△MBE$
$\angle DMC=\angle BME$ (Vertically opposite angles)
$\angle DCM=\angle MBE$ (Alternate angles)
$MC=MB$ (M is mid point of BC)
Thus, $△DMC\cong △EMB$ (ASA rule)
Thus, $DC=BE$ (By CPCT)
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, $\frac{PE}{PD}=\frac{AE}{CD}$
$\frac{PE}{PD}=\frac{AB+BE}{CD}$ (since AB = BE = CD)
$\frac{PE}{PD}=2$
$PE=2PD$