Option : (B)
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 ΔABC,
P and Q are the mid-points of AB and BC respectively.
Therefore,
PQ ‖ AC and PQ = \(\frac{1}{2}\)AC (Mid-point theorem) ..(i)
Similarly,
In ΔADC,
SR ‖ AC and SR = \(\frac{1}{2}\)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 ΔBCD,
Q and R are the mid-points of sides BC and CD respectively
Therefore,
QR ‖ BD and QR = \(\frac{1}{2}\)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 = 90°
Now,
In quadrilateral EHOS, we have
SE || OH
Therefore,
∠AOD + ∠AES = 180°
(Corresponding angle)
∠AES = 180°- 90° = 90°
Again,
∠AES + ∠SEO = 180°(Linear pair)
∠SEO = 180° - 90° = 90°
Similarly,
SH || EO
Therefore,
∠AOD + ∠DHS = 180°
(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°
90° + 90° + 90° + ∠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.