Given in the question,
ABCD is a rhombus and P, Q, R and S are the mid-points of thesides AB, BC, CD and DA respectively.
To Prove,
PQRS is a rectangle.
Construction,
Join AC and BD.
Proof:
In ΔDRS and ΔBPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
∠SDR = ∠QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
, ΔDRS ≅ ΔBPQ[SAS congruency]
RS = PQ [CPCT]———————- (i)
In ΔQCR and ΔSAP,
RC = PA (Halves of the opposite sides of the rhombus)
∠RCQ = ∠PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
, ΔQCR ≅ ΔSAP[SAS congruency]
RQ = SP [CPCT]———————- (ii)
Now,
In ΔCDB,
R and Q are the mid points of CD and BC respectively.
⇒ QR ||BD
also,
P and S are the mid points of AD and AB respectively.
⇒ PS ||BD
⇒ QR ||PS
, PQRS is a parallelogram.
also, ∠PQR =90°
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
,PQRS is a rectangle.