Question

18. 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. Prove that DA = AR and
CQ = QR​

Answer 1

Step-by-step explanation:

You can solve it in the following way.

Answer 2

Answer:

Hence proved DA = AR and  CQ = QR.

Step-by-step explanation:

Given ABCD is a parallelogram

AP║CR

CP = PD

Since, ΔAPD ≈ ΔRCD      (AAA)

⇒$\frac{DA}{DR }= \frac{DP}{DC}$           (CPCT)

⇒$\frac{DA}{DR} = \frac{1}{2}$       (As 'p' is the midpoint of CD)

⇒2DA = DR

⇒ 2DA = DA + AR

⇒ DA = AR

Hence proved DA = AR

Bow, ∠APD = ∠QCD and ∠BQC = ∠APD   (AP║QC)

⇒∠APD = ∠QCD = ∠BQC

and ∠RDC = ∠CBQ    (opposite angles of a parallelogrm is equal)

∠BCQ = ∠CRD    (properties of parallelogrm)

⇒ΔBCQ ≈ ΔDCR      (AAA)

Now ΔAPD ≈ ΔCBQ

PD = BQ

or $\frac{CD}{2} = BQ$     ('p' ia the mid point of CD)

or $\frac{AB}{2} = BQ$     (parallelogrm AB = CD)

∴Q is the mid point of AB

Again, as ΔBQC ≈ ΔDCR

$\frac{BQ}{CQ} = \frac{CD}{CR}$

⇒2CQ = CR ( CD = 2BQ)

⇒2CQ = CQ + QR

⇒CQ = QR.

Hence proved CQ = QR.