Question

Show that the Quadrilateral formed by joining the mid point of the sides of a square is a square
​

Answer 1

$\huge\star\mathcal\green{ᴀɴsᴡᴇʀ}$

In a square ABCD, P,Q,R and S are the mid-points of AB,BC,CD and DA respectively.

⇒ AB=BC=CD=AD [ Sides of square are equal ]

In △ADC,

SR∥AC and SR=

2

1

AC [ By mid-point theorem ] ---- ( 1 )

In △ABC,

PQ∥AC and PQ=

2

1

AC [ By mid-point theorem ] ---- ( 2 )

From equation ( 1 ) and ( 2 ),

SR∥PQ and SR=PQ=

2

1

AC ---- ( 3 )

Similarly, SP∥BD and BD∥RQ

∴ SP∥RQ and SP=

2

1

BD

and RQ=

2

1

BD

∴ SP=RQ=

2

1

BD

Since, diagonals of a square bisect each other at right angle.

∴ AC=BD

⇒ SP=RQ=

2

1

AC ----- ( 4 )

From ( 3 ) and ( 4 )

SR=PQ=SP=RQ

We know that the diagonals of a square bisect each other at right angles.

∠EOF=90

o

.

Now, RQ∥DB

RE∥FO

Also, SR∥AC

⇒ FR∥OE

∴ OERF is a parallelogram.

So, ∠FRE=∠EOF=90

o

(Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R=90

o

and SR=PQ=SP=RQ

∴ PQRS is a square.

Answer 2

Answer:

ᴀɴsᴡᴇʀ

In a square ABCD, P,Q,R and S are the mid-points of AB,BC,CD and DA respectively.

⇒ AB=BC=CD=AD [ Sides of square are equal ]

In △ADC,

SR∥AC and SR=

2

1

AC [ By mid-point theorem ] ---- ( 1 )

In △ABC,

PQ∥AC and PQ=

2

1

AC [ By mid-point theorem ] ---- ( 2 )

From equation ( 1 ) and ( 2 ),

SR∥PQ and SR=PQ=

2

1

AC ---- ( 3 )

Similarly, SP∥BD and BD∥RQ

∴ SP∥RQ and SP=

2

1

BD

and RQ=

2

1

BD

∴ SP=RQ=

2

1

BD

Since, diagonals of a square bisect each other at right angle.

∴ AC=BD

⇒ SP=RQ=

2

1

AC ----- ( 4 )

From ( 3 ) and ( 4 )

SR=PQ=SP=RQ

We know that the diagonals of a square bisect each other at right angles.

∠EOF=90

o

.

Now, RQ∥DB

RE∥FO

Also, SR∥AC

⇒ FR∥OE

∴ OERF is a parallelogram.

So, ∠FRE=∠EOF=90

o

(Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R=90

o

and SR=PQ=SP=RQ

∴ PQRS is a square.