Question

ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the
mid-points of the sides, in order, is a rectangle.

Answer 1

Given,
A kite ABCD having AB = AD and BC = CD. P,Q,R, S are the midpoint of sides
AB,BC,CD andDArespectively PQ,QR,RS and spare joined
To prove:
PQRS is a rectangle

Proof:

In ΔABC, P and Q are the midpoints of AB and BC respectively.

∴PQ || AC and PQ = `1/2` AC            ....(i )

In Δ ADC, R and S are the midpoint of CD and AD respectively.

∴ RS || AC and RS = `1/2` AC             .....(ii )

From (i) and (ii), we have

PQ || RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal and parallel. So PQRS is a
parallelogram. Now, we shall prove that one angle of parallelogram PQRS it is a right angle
Since AB = AD

⇒  `1/2` AB = AD  `(1/2)`

⇒  AP = AS                  ...(iii)         [∵ P and S are the midpoints of B and AD respectively]

⇒  `∠`1 = `∠`2                    ....(iv)

Now, in ΔPBQ and ΔSDR, we have

PB = SD                [ ∴AD = AB ⇒ `1/2` AD = `1/2` AB ]

BQ = DR                  ∴ PB = SD

And PQ = SR        [ ∴ PQRS is a parallelogram]

So by SSS criterion of congruence, we have

Δ PBQ ≅  Δ SOR

⇒  `∠`3 = `∠`4         [CPCT ]

Now, `∠`3 + `∠`SPQ + `∠`2 = 180°

 And `∠`1+ `∠`PSR + `∠`4 =180°

∴ `∠`3 + `∠`SPQ + `∠`2 = `∠`1+ `∠`PSR + `∠`4

⇒ `∠`SPQ = `∠`PSR          (`∠`1 = `∠`2 and `∠`3 = `∠`4)

Now,   transversal   PS   cuts   parallel   lines   SR   and   PQ   at   S   and   P   respectively.

∴ `∠`SPQ + `∠`PSR = 180°

⇒ 2`∠`SPQ = 180° = `∠`SPQ = 90°          [ ∵ `∠`PSR = `∠`SPQ]

Thus, PQRS is a parallelogram such that `∠`SPQ = 90°

Hence, PQRS is a parallelogram.