Option : (B)
Given :
ABCD is a rectangle and P, Q, R, S are their midpoints
To Prove :
PQRS is a rhombus
Proof :
In ΔABC,
P and Q are the mid points
So,
PQ is parallel AC
And,
PQ = \(\frac{1}{2}\)AC
(The line segment joining the mid points of 2 sides of the triangle is parallel to the third side and half of the third side)
Similarly,
RS is parallel AC
And,
RS = \(\frac{1}{2}\)AC
Hence,
Both PQ and RS are parallel to AC and equal to \(\frac{1}{2}\)AC.
Hence,
PQRS is a parallelogram
In triangles APS & BPQ,
AP = BP
(P is the mid-point of side AB)
∠PAS = ∠PBQ (90° each)
AS = BQ
(S and Q are the mid points of AD and BC respectively and since opposite sides of a rectangle are equal, so their halves will also be equal)
ΔAPS ≅ ΔBPQ
(By SAS congruence rule)
PS = PQ (By c.p.c.t.)
PQRS is a parallelogram in which adjacent sides are equal.
Hence,
PQRS is a rhombus.