Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square.

Let ABCD be a square in which E, F, G and H are midpoints of AB, BC, CD and DA respectively.

Join EF, FG, GH and HE.

Join AC and BD.

In △ACD,

G and H are mid-points of CD and AD respectively,

∴ GH || AC and GH = $\dfrac{1}{2}$AC …….(i)

In △ABC,

E and F are mid-points of AB and BC respectively,

∴ EF || AC and EF = $\dfrac{1}{2}$AC …….(ii)

Using (i) and (ii) we get,

EF || GH and EF = GH = $\dfrac{1}{2}$AC ……..(1)

In △ABD,

E and H are mid-points of AB and AD respectively,

∴ EH || BD and EH = $\dfrac{1}{2}$BD …….(iii)

In △BCD,

G and F are mid-points of CD and BC respectively,

∴ FG || BD and FG = $\dfrac{1}{2}$BD …….(iv)

Using (iii) and (iv) we get,

EH || FG and EH = FG = $\dfrac{1}{2}$BD ……..(2)

We know that diagonals of square are equal,

AC = BD

Dividing both sides by 2 we get,

$\dfrac{\text{AC}}{2} = \dfrac{\text{BD}}{2}$

Substituting above value in 1 and 2 we get,

EF = GH = EH = FG ………(v)

∴ EFGH is a parallelogram.

In △GOH and △GOF,

OH = OF as diagonals of parallelogram bisect each other.

OG = OG (Common)

GH = GF (From (v))

∴ △GOH ≅ △GOF (SSS axiom of congruency)

∠GOH = ∠GOF (c.p.c.t.c.)

From figure,

⇒ ∠GOH + ∠GOF = 180°

⇒ ∠GOH + ∠GOH = 180°

⇒ 2∠GOH = 180°

⇒ ∠GOH = 90°.

So, the diagonals of EFGH bisect and are perpendicular to each other.

∴ EFGH is a square.

Hence, proved that quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square.