Question

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

Answer 2

Hello Mate!

Given : ABCD is rhombus where PQRS are formed by joining mid points.

To prove : PQRS is a rectangle.

To construct : Join AC and BD.

Proof : Since RQ are formed by joining mid points so, QR || BD and QR = ½ BD _(i)

Since PS are formed by joining mid points so, PS || BD and PS = ½ BD _(ii)

From (i) and (ii) we get,

QR || PS and QR = PS

Hence a pair if side is equal and parallel so PQRS is ||gm.

Now, since QR || BD so MR || ON __(iii)

Similarly, NR || AC __(iv) [ Because SR || AC, since SR is formed by joining mid points ]

From (iii) and (iv) we get, MONR is a ||gm.

As we know that diagonals of rhombus bisect at 90° so < MON = 90°

Now, since MONR is ||gm so its opposite angles will be equal. So,

< MRN = 90°

Or < QRS = 90°

Hence PQRS is such ||gm whose one if the angle ( < QRS ) is 90° so PQRS is a rectangle.

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ.