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Prove that the quadrilateral formed by joining the midpoints of consecutive sides of rectangle is a rhombus and PLEASE PROVE THE VICE-VERSA ALSO.

To show that the quadrilateral formed by joining the mid point of the pair of adjacent sides of rectangle is a rhombus. 

Here is the proof for the mentioned result:

Let ABCD is a rectangle such as AB = CD and BC = DA. P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.
However, the diagonals of a rectangle are equal.
∴ AC = BD … (5)
By using equation (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus.

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To prove that the quadrilateral formed by joining the mid points of sides of a rhombus is a rectangle..

ABCD is a rhombus. P, Q , R and S are the mid-points of the sides AB, BC, CD and DA respectively.

Join AC.

In ΔABC, P and Q are the mid points of AB and BC respectively

∴ PQ || AC and PQ =  AC ... (1) (Mid point theorem).

Similarly,

RS || AC and RS =  AC ... (2) (Mid point theorem).

From (1) and (2), we get

PQ || RS and PQ = RS.

Thus, PQRS is a parallelogram (A quadrilateral is a parallelogram, if one pair of opposite sides is parallel and equal)

AB = BC (Given)

⇒ PB = BQ (P and Q are mid points of AB and BC respectively)

In ΔPBQ,

PB = BQ

∴ ∠BQP = ∠BPQ ... (3) (Equal sides have equal angles opposite to them)

In ΔAPS and ΔCQR,

AP = CQ 

AS = CR 

PS = RQ (Opposite sides of parallelogram are equal)

∴ ΔAPSΔCQR (SSS congruence criterion)

⇒ ∠APS = ∠CQR ... (4) (CPCT)

Now,

∠BPQ + ∠SPQ + ∠APS = 180°

∠BQP + ∠PQR + ∠CQR = 180°

∴ ∠BPQ + ∠SPQ + ∠APS = ∠BQP + ∠PQR + ∠CQR

⇒ ∠SPQ = ∠PQR ... (5) (from (3) and (4))

PS || QR and PQ is the transversal,

∴ ∠SPQ + ∠PQR = 180° (Sum of adjacent interior an angles is 180°)

⇒∠SPQ + ∠SPQ = 180° (from (5))

⇒ 2 ∠SPQ = 180°

⇒ ∠SPQ = 90°

Thus, PQRS is a parallelogram such that ∠SPQ = 90°.

Hence, PQRS is a rectangle.

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