- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# $ \mathrm{ABCD} $ is a rhombus and $ \mathrm{P}, \mathrm{Q}, \mathrm{R} $ and $ \mathrm{S} $ are the mid-points of the sides $ \mathrm{AB}, \mathrm{BC}, \mathrm{CD} $ and DA respectively. Show that the quadrilateral $ \mathrm{PQRS} $ is a rectangle.

**Given:**

\( \mathrm{ABCD} \) is a rhombus and \( \mathrm{P}, \mathrm{Q}, \mathrm{R} \) and \( \mathrm{S} \) are the mid-points of the sides \( \mathrm{AB}, \mathrm{BC}, \mathrm{CD} \) and DA respectively.

**To do:**We have to show that the quadrilateral \( \mathrm{PQRS} \) is a rectangle.

**Solution:**

Join $PQ,QR,RS,PS,AC$ and $BD$.

In $\triangle DRS$ and $\triangle BPQ$,

$DS = BQ$ ($\frac{AD}{2}=\frac{BC}{2}$)

$\angle SDR = \angle QBP$ (Opposite angles of a rhombus are equal to each other)

$DR = BP$ ($\frac{CD}{2}=\frac{AB}{2}$)

Therefore, by SAS congruency, we get,

$\triangle DRS \cong \triangle BPQ$

This implies,

$RS = PQ$ (CPCT)............(i)

$In \triangle CQR$ and $\triangle ASP$,

$RC = PA$ ($\frac{CD}{2}=\frac{AB}{2}$)

$\angle RCQ = \angle PAS$ (Opposite angles of the rhombus)

$CQ = AS$ ($\frac{BC}{2}=\frac{AD}{2}$)

Therefore, by SAS congruency, we get,

$\triangle QCR \cong \triangle SAP$

This implies,

$RQ = SP$ (CPCT)...............(ii)

In $\triangle CBD$,

$R$ and $Q$ are the mid points of $CD$ and $BC$ respectively.

This implies,

$QR \| BD$

In $\triangle ABD$,

$P$ and $S$ are the mid points of $AD$ and $AB$ respectively.

This implies,

$PS \| BD$

Therefore,

$QR \| PS$............(iii)

From (i), (ii) and (iii), we get,

$PQRS$ is a parallelogram.

$AB$ is a straight line.

$\angle APS + \angle SPQ + \angle QPB = 180^o$

$BC$ is a straight line.

$\angle PQB + \angle PQR + \angle CQR = 180^o$

$\angle APS + \angle SPQ + \angle QPB = \angle PQB + \angle PQR + \angle CQR$

$\angle APS + \angle SPQ + \angle QPB = \angle QPB + \angle PQR + \angle APS$

$\angle SPQ = \angle PQR$

$\angle SPQ + \angle PQR = 180^o$ (Adjacent angles of a parallelogram are supplementary)

$2\angle PQR = 180^o$

$PQR = 90^o$

In $PQRS$,

$RS = PQ$

$RQ = SP$

$\angle Q = 90^o$

Therefore,

$PQRS$ is a rectangle.

- Related Questions & Answers
- Maximize the number of segments of length p, q and r in C++
- Count pairs (p, q) such that p occurs in array at least q times and q occurs at least p times in C++
- Program to calculate area and perimeter of a rhombus whose diagonals are given What is rhombus in C++?
- Find all pairs (a,b) and (c,d) in array which satisfy ab = cd in C++
- Find the canonical cover of FD {A->BC, B->AC, C->AB} in DBMS
- Difference between CD and DVD
- What according to you are the good and bad sides of the #MeToo movement?
- Why are buttons on men’s and women’s clothes placed on different sides?
- What are the P class and NP class in TOC?
- Difference between ++*p, *p++ and *++p in c++
- Difference between ++*p, *p++ and *++p in C
- What are the differences between Non-persistent and p-Persistent CSMA?
- Find smallest number K such that K % p = 0 and q % K = 0 in C++
- Difference between CSMA/CA and CSMA/CD
- Minimum positive integer value possible of X for given A and B in X = P*A + Q*B in C++
- Evaluate a polynomial at points x and x is broadcast over the columns of r for the evaluation in Python