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show that the quadrilateral formed by joining the mid - points of consecutive sides of a square is also a square.




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Answer 1

$\sf\huge\underline\pink {Answer:-}$

• Let a square ABCD in which L,M,N&O
• are the midpoints .

in △AML and △CNO

• AM = CN.

( AB = DC and M and O are the midpoints )

• AL = CM

( AD = BC and L and N are the midpoints )

• ∠MAL = angle NCO

( all angles of a square = 90 degree )

by AAS critaria

• △AML CONGRUENT to △CNO

∴ ML = ON ⠀⠀⠀⠀⠀⠀( CPCT )

similarly,

• In △MBN CONGRUENT to LDO

• △AML is CONGRUENT to triangle

now ,

• in △AML ,

∠ AML = ∠ ALM ( AM = AL )

= 45 degree

similarly

• in △LDO

∠ DLO = 45 degree

∴ ⠀⠀ ⠀∠ MLO = 90 degree

By the properties of SQUARE

All sides are equal and angles are 90 degree .

Answer 2

$\sf\large\underline\green{AnSwEr}$

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=½ AC [ By mid-point theorem ] ---- ( 1 )

In △ABC,

PQ ∥ AC and PQ=½AC [ By mid-point theorem ] ---- ( 2 )

From equation ( 1 ) and ( 2 ),

SR ∥ PQ and SR=PQ=½ AC ---- ( 3 )

Similarly, SP ∥ BD and BD ∥ RQ

∴ SP ∥ RQ and SP=½ BD

and RQ= ½BD

∴ SP=RQ=½BD

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

∴ AC=BD

⇒ SP=RQ= ½ 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°

.

Now, RQ∥DB

RE∥FO

Also, SR∥AC

⇒ FR∥OE

∴ OERF is a parallelogram.

So, ∠FRE=∠EOF=90°

(Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R=90°

and SR=PQ=SP=RQ

∴ PQRS is a square.