Question

P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Then:

• A. DA = AR and CQ = QR
• B. PA = AR
• C. CR = DR
• D. CR = 2CQ

Answer 1

Answer: (A), (D)

The correct options are
A

DA = AR and CQ = QR

D

CR = 2CQ

Given in a parallelogram ABCD, P is the midpoint of DC.

ABCD is a parallelogram.


$\therefore$ BC=AD and BC||AD

Also, DC=AB and DC||AB

Since P is the mid-point of DC.

$\therefore$ $DP=PC=\frac{1}{2}DC$

Now, QC ∥AP and PC ∥ AQ.

So, APCQ is a parallelogram,

$AQ=PC=\frac{1}{2}DC$

$=\frac{1}{2}AB=BQ$ [$\because DC=AB$] . . . . . .(i)

Now, in $\Delta AQR$ and $\Delta BQC$, AQ = BQ [from Eq. (i)]

$\angle AQR=\angle BQC$ [vertically opposite angles]

and $\angle ARQ=\angle BCQ$ [Alternate interior angles]

$\therefore$ $\Delta AQR=\Delta BQC$ [by AAS congruence rule]

$\therefore$ AR = BC [by CPCT rule]

But, BC = DA

$\therefore$ AR = DA

Also, CQ = QR [by CPCT rule]