Q. 134.0( 4 Votes )

# The figure formed

Answer :

Let ABCD is a square such that AB = BC = CD = DA, AC = BD and P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.

In P and Q are the mid-points of AB and BC respectively.

Therefore,

PQ AC and PQ = AC (Mid-point theorem) (i)

Similarly,

In SR AC and SR = AC (Mid-point theorem) (ii)

Clearly,

PQ SR and PQ = SR

Since, in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other. Hence, it is a parallelogram.

Therefore,

PS QR and PS = QR (Opposite sides of a parallelogram) (iii)

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

Therefore,

QR BD and QR = BD (Mid-point theorem) (iv)

However, the diagonals of a square are equal

Therefore,

AC = BD (v)

By using equation (i), (ii), (iii), (iv) and (v), we obtain

PQ = QR = SR = PS

We know that, diagonals of a square are perpendicular bisector of each other

Therefore,

AOD = AOB = COD = BOC = 90o

Now, in quadrilateral EHOS, we have

SE || OH

Therefore,

AOD + AES = 180o (Corresponding angle)

AES = 180° - 90°

= 90°

Again,

AES + SEO = 180o (Linear pair)

SEO = 180° - 90°

= 90°

Similarly,

SH || EO

Therefore,

AOD + DHS = 180o (Corresponding angle)

DHS = 180° - 90° = 90°

Again,

DHS + SHO = 180° (Linear pair)

SHO = 180° - 90°

= 90°

Again,

In quadrilateral EHOS, we have

SEO = SHO = EOH = 90°

Therefore, by angle sum property of quadrilateral in EHOS, we get

SEO + SHO + EOH + ESH = 360°

90o + 90o + 90o + ESH = 360°

ESH = 90°

In the same manner, in quadrilateral EFOP, FGOQ, GHOR, we get

HRG = FQG = EPF = 90°

Therefore, in quadrilateral PQRS, we have

PQ = QR = SR = PS and ESH = HRG = FQG = EPF = 90°

Hence, PQRS is a square.

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