Question

ABCD is a rhombus and P Q R and S are the midpoints of the sides ab BC CD and D A respectively show that the quadrilateral pqrs is a rectangle

Answer 1

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

To Prove-PQRS is a rectangle

Construction,
AC and BD are joined.

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)
Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.
RS = PQ by 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)
Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.
RQ = SP by 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
Thus, PQRS is a parallelogram.
also, ∠PQR = 90° 
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
Thus, PQRS is a rectangle.

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Answer 2

Answer:

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