Question

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square.

Answer 1

Given: In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.

Also, AC = BD and AC ⊥ BD.

To prove: PQRS is a square.

Proof: Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

SR || AC and SR = `1/2` AC   ...(i)

In ΔABC, P and Q are the mid-points of AB and BC, then by mid-point theorem,

PQ || AC and PQ = `1/2` AC  ...(ii)

From equations (i) and (ii),

PQ || SR and PQ = SR = `1/2` AC  ...(iii)

Similarly, in ΔABD, by mid-point theorem,

SP || BD and SP = `1/2` BD = `1/2` AC  [Given, AC = BD] ...(iv)

And ΔBCD, by mid-point theorem,

RQ || BD and RQ = `1/2` BD = `1/2` AC  [Given, BD = AC] ...(v)

From equations (iv) and (v),

SP = RQ = `1/2` AC  ...(vi)

From equations (iii) and (vi),

PQ = SR = SP = RQ

Thus, all four sides are equal.

Now, in quadrilateral OERF,

OE || FR and OF || ER

∴ ∠EOF = ∠ERF = 90°  ...[∵ AC ⊥ DB ⇒ ∠DOC = ∠EOF = 90° as opposite angles of a parallelogram]

∴ ∠QRS = 90°

Similarly, ∠RQS = 90°

So, PQRS is a square.

Hence proved.