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

By mid-point theorem,

The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.

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

We know that diagonals of a square are equal and bisect each other.

AC = BD

AO = OC = BO = OD

Join EF, FG, GH and HE.

Join AC and BD.

In △ACD,

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

By mid-point theorem,

∴ GH || AC and GH = $\dfrac{1}{2}$ AC

GH = AO (∵ O is the mid-point of AC) …(1)

In △ABC,

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

∴ EF || AC and EF = $\dfrac{1}{2}$ AC

EF = AO (∵ O is the mid-point of AC) …(2)

In △ABD,

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

∴ EH || BD and EH = $\dfrac{1}{2}$ BD

EH = BO (∵ O is the mid-point of BD)

∴ EH = AO …(3)

In △BCD,

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

∴ FG || BD and FG = $\dfrac{1}{2}$ BD

FG = BO (∵ O is the mid-point of BD)

∴ FG = AO ….(4)

From eq.(1), (2), (3) and (4), we get:

EH || FG, EF || GH and EH = FG = GH = EF

Since, both the opposite sides of a quadrilateral are parallel.

∴ EFGH is a parallelogram.

In △GOH and △GOF,

⇒ OH = OF (Diagonals of parallelogram bisect each other)

⇒ OG = OG (Common side)

⇒ GH = GF (Proved above)

∴ △GOH ≅ △GOF (S.S.S axiom)

⇒ ∠GOH = ∠GOF (Corresponding parts of congruent triangles are equal)

From figure,

⇒ ∠GOH + ∠GOF = 180°

⇒ ∠GOH + ∠GOH = 180°

⇒ 2∠GOH = 180°

⇒ ∠GOH = $\dfrac{180°}{2}$

⇒ ∠GOH = 90°

So, the diagonals of EFGH bisect and are perpendicular to each other and all sides of quadrilateral EFGH are equal.

∴ EFGH is a square.

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