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

Name: P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.
Uploaded: Thu, 2021-01-07 22:45
Description: P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.

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

Answers (1)

Given: ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA.

To prove : PQRS is a rhombus.
Proof :
To $\triangle ADC$ , S and R are the mid-points of AD and DC respectively.
By mid-point theorem we get
$SR\parallel AC$    and    $SR=\frac{1}{2}AC$   …..(1)
In $\triangle ABC$ , P and Q are the mid-points of AB and BC respectively.
Then by the mid-point theorem we get
$PQ\parallel AC$    and     $PQ=\frac{1}{2}AC$ …..(2)
From equations 1 and 2, we get
$SR=PQ=\frac{1}{2}AC$ …..(3)
Similarly   $\triangle BCD$ we get
$RQ\parallel BD$ and $RQ=\frac{1}{2}BD$ …..(4)
And in $\triangle BAD$
$SP\parallel BD$   and $SP=\frac{1}{2}BD$ …..(5)
Using equations 4 and 5 we get
$RQ=SP=\frac{1}{2}BD$
It is given that AC = BD
$RQ=SP=\frac{1}{2}AC$   ......(6)
Now equate equations 3 and 6 we get
$SR = PQ = SP = RQ$
It shows that all sides of a quadrilateral PQRS are equal.
Hence PQRS is a rhombus.
Hence Proved.