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

Name: P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that AC BD. Prove that PQRS is a rectangle.
Uploaded: Thu, 2021-01-07 23:02
Description: P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that AC BD. Prove that PQRS is a rectangle.

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that $AC \perp BD$ . Prove that PQRS is a rectangle.

Answers (1)

Solution.
Given: ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA and $AC \perp BD$ . To prove: PQRS is a rectangle
It is given that $AC \perp BD$
\ $\angle COD =\angle AOD = \angle AOB = \angle COB = 90^{\circ}$

In $\triangle ADC$ , S and R are the mid-points of AD and DC so by mid-point theorem.
$SR \parallel AC$    and $SR=\frac{1}{2}AC$    …..(1)
Similarly in $\triangle ABC$ ,
$PQ \parallel AC$    & $PQ =\frac{1}{2}AC$ …..(2)
Using 1 and 2
$SR=PQ=\frac{1}{2}AC$ …..(3)
Similarly, $SP\parallel RQ$    & $SP=RQ=\frac{1}{2}BD$   ……(4)
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}
$\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}$
$90^{\circ} + 90^{\circ} + \angle RSP + \angle RQP = 360^{\circ}$
$\angle RSP + \angle RQP = 180^{\circ}$
$\angle RSP =\angle RQP = 90^{\circ}$
If all the angles in parallelogram are $90^{\circ}$ then that parallelogram is a rectangle.
So, PQRS is a rectangle.
Hence Proved