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:
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.
In ΔABC, P and Q are the mid points of AB and BC respectively
∴ PQ || AC and PQ = AC ... (1) (Mid point theorem).
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)
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)
∠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.