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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