Question

ABCD is a rhombus in which P, Q, R and S are the mid point of the sides AB, BC, CD and DA respectively. Show that quadrilateral PQRS is a rectangle.​

Answer 1

$\underline{ \sf \large{Solution- }}$

➢ Diagram : See the attachment

In △ADC,

• S is mid point of AD
• R is mid point of DC

By Mid Point Theorm,

$\sf \implies SR \parallel AC$

$\sf \implies SR = \dfrac{1}{2} AC \: \: \: \: \: \: \: \: \: \: \: \bf[1]$

Now,

In △ABC,

• P is mid point of AB
• Q is mid point of BC

By Mid Point Theorm,

$\sf \implies PQ\parallel AC$

$\sf \implies PQ = \dfrac{1}{2} AC \: \: \: \: \: \: \: \: \: \: \: \bf[2]$

From [1] and [2]

$\sf \implies SR \parallel PQ \: and \: SR = PQ$

Now,

➢ If one pair of opposite side of a quadrilateral is equal and parallel then it is said to be parallelogram.

• PQRS is a parallelogram.

Now,

$\sf \implies PS \parallel BD \: \: \: [Mid \: point \: Th^m]$

$\sf \implies PM \parallel N O \: \: \: [Parts \: of \parallel lines]\: - - - - \bf[1]$

Now,

$\sf \implies PQ \parallel AC \: \: \: [Mid \: point \: Th^m]$

$\sf \implies PN \parallel MO \: \: \: [Parts \: of \parallel lines]\: - - - - \bf[2]$

From [1] and [2]

• PMON is a parallelogram.

$\sf \implies \angle1 = \angle2 \: \: [Opposite \: sides \: of \: a \parallel ^{gm } \: are \: equal]$

So,

$\sf \implies \angle1 = 90^o \: \: [Diagonals \: of \: a \: rhombus \: bisect \: at \: 90^o ]$

$\sf \implies \angle2 = 90^o$

➢ Parallelogram with one corner angle 90° is a rectangle.

Answer 2

Answer:

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.

Step-by-step explanation: