In the adjoining figure, ABCD and PQBA are two parallelograms. Prove that :

(i) DPQC is a parallelogram.

(ii) DP = CQ.

(iii) ΔDAP ≅ ΔCBQ.

(i) Given,

ABCD and PQBA are two parallelograms.

DC ∥ AB ….(1)

AB ∥ PQ …..(2)

From (1) and (2) we have,

∴ DC ∥ PQ

Opposite sides of a parallelogram are equal.

Thus, in //gm ABCD

DC = AB …..(3)

Thus, in //gm PQBA

AB = PQ ……(4)

From (3) and (4) we have,

∴ DC = PQ

Since, the pair of opposite sides DC and PQ are equal and parallel. Therefore, DPQC is a parallelogram.

Hence, proved that DPQC is a parallelogram.

(ii) We know that,

DPCQ is a parallelogram.

Opposite sides of a parallelogram are equal.

∴ DP = CQ

Hence, proved that DP = CQ.

(iii) In triangle DAP and CBQ,

DA = CB [opposite sides of parallelogram ABCD]

AP = BQ [opposite sides of parallelogram PQBA]

DP = CQ [opposite sides of parallelogram DPQC]

∴ ΔDAP ≅ ΔCBQ [By SSS rule]

Hence, proved that DP = CQ.