Question

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PQRS is a parallelogram. PS is produced to meet M so that SM = SR and MR is produced to meet PQ produced at N. Prove that QN= QR.

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

PQRS is a parallelogram. PS is produced to meet M so that SM = SR and MR is produced to meet PQ produced at N.

In triangle SMR

As it is given that, SM = SR

We know, angle opposite to equal sides are always equal.

So, using this result, we get

∠SMR = ∠SRM = x (say)

Now, again in triangle SMR

We know, Exterior angle is equals to sum of interior opposite angles.

So, ∠PSR = ∠SMR + ∠SRM = x + x = 2x

Now, as PQRS is a parallelogram.

We know, opposite angles of a parallelogram are equal.

Therefore, ∠PSR = ∠RQP = 2x

Now, PQ || RS

∠SRQ + 2x = 180°

So, ∠SRQ = 180° - 2x

Now, MRN is a straight line.

∠QRN + x + 180° - 2x = 180°

So, ∠QRN = x

Now, In triangle QRN

We know, Exterior angle of a triangle is equals to sum of interior opposite angles.

So, using this, we get

∠RQP = ∠QRN + ∠QNR

2x = x + ∠QNR

This implies, ∠QNR = x

$\rm\implies \: \angle \: QRN \: = \: \angle \: QRN \: = \: x$

We know,

Side opposite to equal angles are equal.

So, using this result, we get

$\rm\implies \:QR \: = \: QN$

Hence, Proved

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

1. Sum of interior angles of a triangle is supplementary.

2. Sum of exterior angles of a triangle is 360°.

3. Angle opposite to longest side is always greater.

4. Side opposite to greater angle is always longest.

Answer 2

In △SMR,

⇒  SM=SR               [ Given ]

∴  ∠SMR=∠SRM         [ Angles opposite to equal sides are equal ]

Let ∠SMR=∠SRM=x     ----- ( 1 )

As ∠PSR is an exterior angle of △SMR

⇒  So, ∠PSR=∠SMR+∠SRM

⇒  ∠PSR=x+x=2x      ---- ( 2 )

⇒  ∠PSR=∠PQR       [ Opposite angles of parallelogram PQRS ]

⇒  So, ∠PQR=2x         ----- ( 3 )

As PM∥QR

⇒  So, ∠PSR+∠QRS=180  

o

.

⇒  2x+∠QRS=180  

o

.

⇒  ∠QRS=180  

o

−2x              ----- ( 4 )

As, ∠QRS+∠SRM+∠QRN=180  

o

 

⇒  (180  

o

−2x)+x+∠QRN=180  

o

         [ From ( 1 ) and ( 4 ) ]

⇒  (180  

o

−x)+∠QRN=180  

o

 

⇒  ∠QRN=x    ---- ( 5 )

Also, ∠PQR is an exterior angle of △QRM

So, ∠PQR=∠QRN+∠QNR

⇒  2x=x+∠QNR            [ From ( 5 ) and ( 3 ) ]

⇒  ∠QNR=x            ---- ( 6 )

In △QNR,

⇒  ∠QRN=∠QNR          [ From ( 5 ) and ( 6 ) ]

∴  QR=QN          [ Angles opposite to sides are equal ]