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Show that the quadrilateral PQRS is a rhombus.

Answer :

Given: ABCD is a rectangle.

\(\Rightarrow \) \(\angle{A}\) = \(\angle{B}\) = \(\angle{C}\) = \(\angle{D}\) = \(90^\circ\) and

AB = CD and BC = AD.

Also, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Therefore, by midpoint theorem,

PQ || BD and PQ = \(\frac{1}{2} \) BD and

SR || AC and SR = \(\frac{1}{2} \) AC

In rectangle ABCD,

AC = BD

\(\therefore \) PQ = SR ...(i)

Now, in \(\triangle{ASP}\) and \(\triangle{BQP}\), we have,

AP = BP ...(Given)

AS = BQ ...(Given)

\(\angle{A}\) = \(\angle{B}\) ...(Given)

\(\therefore \) \(\triangle{ASP}\) \(\displaystyle \cong \) \(\triangle{BQP}\) ...(By SAS rule)

\(\therefore \) SP = BQ ...(ii)(By CPCT)

Similarly, in \(\triangle{RDS}\) and \(\triangle{RCQ}\), we have,

SD = CQ ...(Given)

DR = RC ...(Given)

\(\angle{C}\) = \(\angle{D}\) ...(Given)

\(\therefore \) \(\triangle{RDS}\) \(\displaystyle \cong \) \(\triangle{RCQ}\) ...(By SAS rule)

\(\therefore \) SR = RQ ...(iii)(By CPCT)

Thus, from eq. (i), (ii) and (iii), we can say that,

The quadrilateral PQRS is a rhombus is proved.

- ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show thati) SR || AC and SR = (1/2) AC ii) PQ = SRiii) PQRS is a parallelogram.
- ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
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