Question

ABCD is a rhombus P,Q,R and S are the mid-points of sides AB,BC,CD and AD respectively. Show that 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 ∆ DRS and ∆ BPQ

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 ∆ QCR and ∆ SAP,

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

Given :

• ABCD is a rhombus P, Q, R and S are the mid - points of sides and AD respectively.

To prove :

• Prove PQRS is a reactangle ?

Construction :

• AC and BD are joined.

Proof :

• In ∆DRS and ∆BPQ

DS = BQ $~~$${\sf{[Halves \;of\; the\; opposite \;sides \;of\; the \;rhombus]}}$

$\angle$SDR = $\angle$QBP $~~$${\sf{[Opposite \;angle \;of \;the\; rhombus]}}$

DR = BP $~~$${\sf{[Halves\; of\; the \;opposite \;sides \;of \;the\; rhombus]}}$

Thus,

• ∆DRS ≈ ∆BPQ by SAS congruence condition.

RS = PQ by CPCT $~~~~~$${\sf{(equ...i)}}$

In ∆QCR and ∆SAP,

RC = PA $~~$${\sf{[Halves \;of\; the \;opposite \;sides \;of \;the\; rhombus]}}$

$\angle$ RCQ = $\angle$PAS = $~~$${\sf{[Opposite \;angle \;of \;the \;rhombus]}}$

CQ = AS $~~$${\sf{[Halves \;of \;the \;opposite \;sides \;of \;the \;rhombus]}}$

Thus,

• ∆QCR ≈ ∆SAP by SAS congruence condition.

RQ = SP by CPCT $~~~~~$${\sf{(equ...ii)}}$

Now, we have

• In ∆CDB,

R and Q are the mid - points of CD and BC respectively.

$\rightarrow$QR ll BD

Also, we have

• P and S are the mid - points of AD and AB respectively.

$\rightarrow$PQ ll BD

$\rightarrow$QR ll PS

Thus,

• PQRS is a parallelogram.

Again also, we have

$\angle$PQRS = 90°

Now,

• In PQRS,

RS = PQ and RQ = SP from (i) & (ii)

$\angle$Q = 90°

Thus,

• PQRS is a reactangle.

$~~~~$$\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Proved!}}}}$

$~~~~~~~~~~~~~~~$ ____________________