Question

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

Answer 1

Join AC and BD

In ΔACD, G and H are the mid-points of DC and AC respectively.

Therefore, GH || AC and GH = `(1)/(2)"AC"`   ...(i)

In ΔABC, E and F are the mid-points of AB and BC respectively.

Therefore, EF || AC and EF = `(1)/(2)"AC"`   ...(ii)

From (i) and (ii)

EF || GH and EF = GH = `(1)/(2)"AC"`   ...(iii)

Similarly, it can be proved that

EF || GH and EH = GF = `(1)/(2)"BD"`   ...(iv)

But AC = BD   ...(Diagonals of a square are equal)

Dividing both sides by 2,

`(1)/(2)"BD" = (1)/(2)"AC"`

From (iii) and (iv)

EF = GH = EH = GF

Therefore, EFGH is a parallelogram.

Now in ΔGOH and ΔGOF

OH = OF   ...(Diagonals of a parallelogram bisect each other)

OG = O   ...(Common)

GH = GF

∴ ΔGOH ≅ ΔGOF

∴ ∠GOH = ∠GOF

Now, ∠GOH + ∠GOF = 180°

⇒ ∠GOH + ∠GOH = 180°

⇒ 2∠GOH = 180°

⇒ ∠GOH = 90°

Therefore, diagonals of parallelogram EFGH bisect each other and are perpendicular to each other.

Thus, EFGH is a square.