Question

ABCD is a parallelogram. in which P and Q are midpoints of opposite sides AB and CD /
Fig. 8.45). If AQ intersects DP at S and BQ intersects CP at R, show that:
APCQ is a parallelogram.
DPBQ is a parallelogram.
PSQR is a parallelogram.
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Answer 1

ABCD is a parallelogram. P and Q mid points of AB and CD respectively. AQ intersects DP at S and BQ intersects CP at R.

To prove,

1) APCQ is a parallelogram.

•DC is parallel and equal to AB (ABCD is a parallelogram)

•In quadrilateral APCQ,

QC and AP is parallel and equal(Half of DC and AB).

•Hence, APCQ is a parallelogram.

2) DPBQ is parallelogram.

•In quadrilateral DPBQ,

DQ is equal and parallel to PB (Half of DC and AB).

•hence, DPBQ is a parallelogram.

3)PSQR is a parallelogram.

•In parallelogram APCQ, AQ and PC are equal and parallel.

•In quadrilateral PSQR,

SQ and PR are equal and opposite (half of AQ and PC).

•Hence, PSQR is a parallelogram.

Answer 2

The proof is given below:

If P is the mid point of AB and Q is the mid point of CD then

AP = PB = AB/2

CQ = QD = CD/2

But ABCD is a parallelogram

∴ AB = CD

∴ AP = PB = CQ = QD

Also

∵ AB ║ CD

∴ AP ║ CQ ║ QD

And PB ║ CQ ║ QD

In quadrilateral APCQ

∵ AP ║ CQ

And, AP = CQ

Therefore, APCQ is a parallelogram   (By theorem: If in a quadrilateral the opposite sides are equal and parallel then the quadrilateral is a parallelogram)

In quadrilateral DPBQ

∵ PB ║ QD

And, PB = QD

Therefore, DPBQ is a parallelogram

It can be easliy proved that APQD and PBCQ are parallelograms

AQ, DP and PC, QB are diagonals of the parallelograms APQD and PBCQ

∴ AS = SQ and DS = SP

And PR = RC and BR = RQ

∵ APCQ is a parallelogram

∴ AQ = PC

Therefore, (1/2)AQ = (1/2)PC

or, SQ = PR

And AQ ║ PC

Therefore, (1/2)AQ ║ (1/2)PC

or, SQ ║ PR

∵ SQ = PR

And SQ ║ PR

Therefore, quadrilateral PSQR is a parallelogram     (Hence Proved)

Hope this answer is helpful.

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