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.

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 $AC \perp BD$ . Prove that PQRS is a square.

Answers (1)

Given: ABCD is a parallelogram and P, Q, R and S are the mid-points of sides AB, BC, CD and AD. Also AC = BD and $AC \perp BD$ .
To prove: PQRS is a square.

Proof: In $\triangle ADC$ , S and R are the mid-points of sides AD and DC by mid-point theorem.
$SR \parallel AC$    and    $SR =\frac{1}{2}AC$   …..(1)
In $\triangle ABC$ , P and Q are the mid-points of AB and BC respectively. Therefore, by the mid-point theorem
$PQ\parallel AC$ and    $PQ=\frac{1}{2}AC$   …..(2)
From equations 1 and 2, we get
$PQ\parallel SR$    and    $PQ=\frac{1}{2}AC$ …..(3)
Similarly, in $\triangle BCD$ , $RQ\parallel BD$ and R, Q are midpoints of CD, CB respectively, Therefore, by mid-point theorem
$RQ=\frac{1}{2}BD=\frac{1}{2}AC$ {Given BD = AC} …..(4)
And in $\triangle ABD$ , $SP\parallel BD$ and S, P is midpoints of AD, and AB respectively. Therefore, by the mid-point theorem
$SP=\frac{1}{2}BD=\frac{1}{2}AC$   {Given AC = BD  …..(5)
From equations 4 and 5 we get
$SP=RQ=\frac{1}{2}AC$ …..(6)
From equations 3 and 6 we get
$PQ = SR = SP = RQ$
Thus, all sides are equal.
In quadrilateral EOFR,
$OE \parallel FR, OF \parallel ER$
$\therefore \angle EOF =\angle ERF = 90^{\circ}$
{ $\angle COD = 90^{\circ}$ because $AC \perp BD$ and opposite angles of a parallelogram are equal}
In quadrilateral RSPQ
$\angle SRQ = \angle ERF = \angle SPQ = 90^{\circ}$   {Opposite angles in a parallelogram are equal}
$\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}$
$90^{\circ} + 90^{\circ}+ \angle RSP +\angle RQP = 360^{\circ}$
$\angle RSP +\angle RQP = 180^{\circ}$
$\angle RSP = \angle RQP = 90^{\circ}$
All the sides are equal and all the interior angles of the quadrilateral are $90^{\circ}$
Hence, PQRS is a square.
Hence Proved