Solution:

\(\Box\)ABCD is a parallelogram. ... (Given)
∴ AB | | CD ... (Opposite sides of a parallelogram)
∴ AP | | CQ
  ... (∵ A – P – B and C – Q – D) ... (i)

Also, AB = CD ... (Opposite sides of a parallelogram)

∴ \(\displaystyle \frac{1}{2}\)AB = \(\displaystyle \frac{1}{2}\)CD

∴ AP = CQ
  ... (∵ P and Q are midpoints of AB and CD respectively) ... (ii)

A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent. ... (Theorem)

∴ From (i) and (ii):
∴ \(\Box\)APCQ is a parallelogram.

Solution:

Given:

\(\Box\)ABCD is a rectangle.

To Prove:

\(\Box\)ABCD is a parallelogram.

Proof:

 \(\Box\)ABCD is a rectangle. ... (Given)

∴ \(\angle\)A = \(\angle\)B = \(\angle\)C = \(\angle\)D = 90°
  ... (Angles of a rectangle)

∴ \(\angle\)A \(\cong\) \(\angle\)C and \(\angle\)B \(\cong\) \(\angle\)D

A quadrilateral is a parallelogram if its pairs of opposite angles are congruent.

∴ \(\Box\)ABCD is a parallelogram.

Proof:

Take point J as shown.

Now, DJ is a median and G is the centroid in \(\triangle\)DEF.
The centroid of a triangle divides the medians in the ratio 2 ∶ 1. ... (Theorem)
∴ DG = 2GJ

But, DG = GH ... (Given)
∴ GH = 2GJ
∴ J is the midpoint of diagonal GH. ... (i)

Also, DJ is a median in \(\triangle\)DEF.
∴ J is the midpoint of diagonal EF. ... (ii)

A quadrilateral is a parallelogram if its diagonals bisect each other. ... (Theorem)
From (i) and (ii),
\(\Box\)GEHF is a parallelogram.

Solution:

Given:

\(\Box\)ABCD is a parallelogram.
in which AD > AB
The bisectors of its angles meet in points P, Q, R and S as shown.

To prove:

\(\Box\)PQRS is a rectangle.

Proof:

\(\Box\)ABCD is a parallelogram. ... (Given)
∴ AD | | BC ... (Opposite sides of a | |^m)
∴ m\(\angle\)DAB + m\(\angle\)ABC = 180°
  ... (Converse of Interior Angles Test)

∴ \(\displaystyle \frac{1}{2} \times m\angle \text{DAB} + \frac{1}{2} \times m\angle \text{ABC} = \frac{1}{2} \times 180^\circ\)

∴ \(\displaystyle m\angle \text{PAB} + m\angle \text{PBA} = 90^\circ\)
  ... (∵ AP and BP are bisectors) ... (i)

Now, in \(\triangle\)APB,
\(\displaystyle m\angle \text{PAB} + m\angle \text{PBA} + m\angle \text{APB} = 180^\circ\) ... (Theorem)
∴ \(\displaystyle 90^\circ + m\angle \text{APB} = 180^\circ\) ... [From (i)]
∴ \(\displaystyle m\angle \text{APB} = 180^\circ - 90^\circ\)
∴ \(\displaystyle m\angle \text{APB} = 90^\circ\)
∴ \(\displaystyle m\angle \text{SPQ} = 90^\circ\) ... (Vertically opposite angles) ... (ii)

Similarly, AB | | DC ... (Opposite sides of a | |^m)
∴ m\(\angle\)CDA + m\(\angle\)DAB = 180°
  ... (Converse of Interior Angles Test)

∴ \(\displaystyle \frac{1}{2} \times m\angle \text{CDA} + \frac{1}{2} \times m\angle \text{DAB} = \frac{1}{2} \times 180^\circ\)

∴ \(\displaystyle m\angle \text{QDA} + m\angle \text{QAD} = 90^\circ\)
  ... (∵ DQ and AQ are bisectors) ... (iii)

Now, in \(\triangle\)DQA,
\(\displaystyle m\angle \text{QDA} + m\angle \text{QAD} + m\angle \text{DQA} = 180^\circ\) ... (Theorem)
∴ \(\displaystyle 90^\circ + m\angle \text{DQA} = 180^\circ\) ... [From (iii)]
∴ \(\displaystyle m\angle \text{DQA} = 180^\circ - 90^\circ\)
∴ \(\displaystyle m\angle \text{DQA} = 90^\circ\)
i.e. \(\displaystyle m\angle \text{PQR} = 90^\circ\) ... (iv)

Also, AD | | BC ... (Opposite sides of a | |^m)
∴ m\(\angle\)BCD + m\(\angle\)CDA = 180°
  ... (Converse of Interior Angles Test)

∴ \(\displaystyle \frac{1}{2} \times m\angle \text{BCD} + \frac{1}{2} \times m\angle \text{CDA} = \frac{1}{2} \times 180^\circ\)

∴ \(\displaystyle m\angle \text{RCD} + m\angle \text{RDC} = 90^\circ\)
  ... (∵ CR and DR are bisectors) ... (v)

Now, in \(\triangle\)CRD,
\(\displaystyle m\angle \text{RCD} + m\angle \text{RDC} + m\angle \text{CRD} = 180^\circ\) ... (Theorem)
∴ \(\displaystyle 90^\circ + m\angle \text{CRD} = 180^\circ\) ... [From (v)]
∴ \(\displaystyle m\angle \text{CRD} = 180^\circ - 90^\circ\)
∴ \(\displaystyle m\angle \text{CRD} = 90^\circ\)
∴ \(\displaystyle m\angle \text{QRS} = 90^\circ\) ... (Vertically opposite angles) ... (vi)

Similarly, AB | | DC ... (Opposite sides of a | |^m)
∴ m\(\angle\)ABC + m\(\angle\)BCD = 180°
  ... (Converse of Interior Angles Test)

∴ \(\displaystyle \frac{1}{2} \times m\angle \text{ABC} + \frac{1}{2} \times m\angle \text{BCD} = \frac{1}{2} \times 180^\circ\)

∴ \(\displaystyle m\angle \text{SBC} + m\angle \text{SCB} = 90^\circ\)
  ... (∵ BS and CS are bisectors) ... (vii)

Now, in \(\triangle\)BSC,
\(\displaystyle m\angle \text{SBC} + m\angle \text{SCB} + m\angle \text{BSC} = 180^\circ\) ... (Theorem)
∴ \(\displaystyle 90^\circ + m\angle \text{BSC} = 180^\circ\) ... [From (vii)]
∴ \(\displaystyle m\angle \text{BSC} = 180^\circ - 90^\circ\)
∴ \(\displaystyle m\angle \text{BSC} = 90^\circ\)
i.e. \(\displaystyle m\angle \text{RSP} = 90^\circ\) ... (viii)

From (ii), (iv), (vi) and (viii):
\(\Box\)PQRS is a rectangle

∴ The bisectors of the four angles of a parallelogram, which is not a rhombus form a rectangle.

Proof:

 AB = CD ... (Opposite sides of a | |^m)
∴ AP + PB = CR + RD
But, AP = CR ... (Given)
∴ PB = RD ... (i)
  ... (Subtracting AP from LHS and CR from RHS)

Also, BC = DA ... (Opposite sides of a | |^m)
∴ BQ + QC = DS + SA
But, BQ = DS ... (Given)
∴ QC = SA ... (ii)
  ... (Subtracting BQ from LHS and DS from RHS)

Now, in \(\triangle\)PBQ and \(\triangle\)RDS,
 seg PB \(\cong\) seg RD ... [From (i)]
 \(\angle\)PBQ = \(\angle\)RDS ... (Opposite angles of a | |^m)
 seg BQ \(\cong\) seg DS ... (Given)
∴ \(\triangle\)PBQ \(\cong\) \(\triangle\)RDS ... (S A S test)
∴ seg PQ \(\cong\) seg RS ... (c s c t) ... (iii)

Also, in \(\triangle\)QCR and \(\triangle\)SAP,
 seg QC \(\cong\) seg SA ... [From (ii)]
 \(\angle\)QCR = \(\angle\)SAP ... (Opposite angles of a | |^m)
 seg CR \(\cong\) seg AP ... (Given)
∴ \(\triangle\)QCR \(\cong\) \(\triangle\)SAP ... (S A S test)
∴ seg QR \(\cong\) seg SP ... (c s c t) ... (iv)

From (iii) and (iv):
\(\Box\)PQRS is a parallelogram.