show that the quadrilateral formed by joining the midpoints of the consecutive side of a square is also a square

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.

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

∴∠AOD = ∠AOB = ∠COD = ∠BOC =  90°

Now, in quadrilateral EHOS, we have 

SE || OH, therefore, ∠AOD + ∠AES = 180°  [corresponding angles]

⇒ ∠AES = 180° - 90° = 90°

Again, ∠AES + ∠SEO = 180° [linear pair]

⇒ ∠SEO  = 180° - 90° = 90°  

Similarly, 

SH || EO, therefore, ∠AOD + ∠DHS = 180° [corresponding angles]

⇒ ∠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.