Question

ABCD is a rhombus and P Q R and S are the mid points of the sides AB BC CD and AD respectively show that PQRS is a triangle
​

Answer 1

Answer:

Step-by-step explanation:

In △ABC, P and Q are the mid- points of AB and BC.  

PQ∥AC and by using mid - point theorem

PQ=  

2

1

​  

AC .

Similarly, in △ADC, R and S are  the mid- points of CD and AD.  

SR∥AC and  by using mid point theorem

SR=  

2

1

​  

AC  

From (1) and (2), we get  

PQ∥RS and PQ=SR

Now, in quadrilateral PQRS its one pair of opposite sides PQ and SR is equal and parallel.  

Therefore,  

PQRS is  a parallelogram  

AB=BC (Sides of a rhombus)

⇒  

2

1

​  

AB=  

2

1

​  

BC

PB=BQ

∠3=∠4

Now, in △APS and △CQR, we have  

AP=CQ (Halves of equal sides AB,BC)

AS=CR (Halves of equal sides AD,CD)

PS=QR (Opp. sides of parallelogram PQRS)

Therefore, △APS≅△CQR ∣ using SSS Congruency Theorem

∠1=∠2 (Corresponding parts of congruent triangles are equal)

Now , ∠1+∠SPQ+∠3=180  

o

 (Linear pair axiom)

Therefore, ∠1+∠SPQ+∠3=∠2+∠PQR+∠4

But, ∠1=∠2 and ∠3=∠4  

Therefore,  

∠SPQ=∠PQR    .....(3)

Since, SP∥RQ and PQ intersects them

Therefore, ∠SPQ+∠PQR=180  

o

....(4) (Since consecutive interior angles are supplementary)

From  (3) and (4), we get  

∠PQR+∠PQR=180  

o

 

2∠PQR=180  

o

 

∠PQR=90  

o

 

∠SPQ=∠PQR=90  

o

 

Thus, PQRS is a parallelogram whose one angle ∠SPQ=90  

o

.

Hence PQRS is a rectangle.