In the adjoining figure, ABCD is a parallelogram. If P and Q are point

1.	In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that AP = 1/3 AD and CQ = 1/3 BC, prove that AQCP is a parallelogram.
Answer» Consider △ ABQ and △ CDP We know that the opposite sides of a parallelogram are equal AB = CD So we get ∠ B = ∠ D We know that DP = AD – PA i.e. DP = 2/3 AD BQ = BC – CQ i.e. BQ = BC – 1/3 BC BQ = (3-1)/3 BC We know that AD = BC So we get BQ = 2/3 BC = 2/3 AD We get BQ = DP By SAS congruence criterion △ ABQ ≅ △ CDP AQ = CP (c. p. c. t) We know that PA = 1/3 AD We know that AD = BC CQ = 1/3 BC = 1/3 AD So we get PA = CQ ∠ QAB = ∠ PCD (c. p. c. t)… (1) We know that ∠ QAP = ∠ A – ∠ QAB Consider equation (1) ∠ A = ∠ C ∠ QAP = ∠ C – ∠ PCD From the figure we know that the alternate interior angles are equal ∠ QAP = ∠ PCQ So we know that AQ and CP are two parallel lines. Therefore, it is proved that PAQC is a parallelogram.

In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that AP = 1/3 AD and CQ = 1/3 BC, prove that AQCP is a parallelogram.

Answer»

Consider △ ABQ and △ CDP

We know that the opposite sides of a parallelogram are equal

AB = CD

So we get ∠ B = ∠ D

We know that

DP = AD – PA

i.e. DP = 2/3 AD

BQ = BC – CQ

i.e. BQ = BC – 1/3 BC

BQ = (3-1)/3 BC

We know that AD = BC

So we get

BQ = 2/3 BC = 2/3 AD

We get BQ = DP

By SAS congruence criterion

△ ABQ ≅ △ CDP

AQ = CP (c. p. c. t)

We know that

PA = 1/3 AD

We know that AD = BC

CQ = 1/3 BC = 1/3 AD

So we get

PA = CQ

∠ QAB = ∠ PCD (c. p. c. t)… (1)

We know that

∠ QAP = ∠ A – ∠ QAB

Consider equation (1)

∠ A = ∠ C

∠ QAP = ∠ C – ∠ PCD

From the figure we know that the alternate interior angles are equal

∠ QAP = ∠ PCQ

So we know that AQ and CP are two parallel lines.

Therefore, it is proved that PAQC is a parallelogram.