Question

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

Answer 1

Answer: Proved QN = QR

Step-by-step explanation:

Given : In ║gm PQRS, SM =SR

To prove QN =QR

Proof :

In Δ SMR

∵ SM = SR

∠SMR =∠SRM

( Angles opposite to equal sides are equal)

Let ∠SMR =∠SRM = X...(1)

As,∠PSR is an exterior angle of ΔSMR

So, ∠PSR = ∠SMR +∠SRM

 ∠PSR = x + x= 2x....(2)

Since ∠PSR =∠PQR (Opposite angles of parallelogram PQRS)

So, ∠PQR = 2x...(3)

As PM║ QR

So, ∠PSR +∠QRS = 180

2x + ∠QRS = 180

∠QRS = 180 - 2x....(4)

As, ∠QRS +∠SRM +∠QRN =180

(180-2x) + x+ ∠QRN = 180  ( from 1 and 4)

(180 - x) + ∠QRN = 180

∠QRN = x ....(5)

Also, ∠PQR is an exterior angle of Δ QRM

So, ∠PQR =∠QRN +∠QNR

 2x = x + ∠QNR (from 5 and 3 )

∠QNR = x....(6)

In Δ QNR,

∠QRN =∠ QNR

(from 5 and 6)

∴ QR = QN ( Angles opposite to sides are equal)

Answer 2

Given that, PQRS is a parallelogram and line SR =SM

To Prove - Line QR = QN

Proof,

∠SRM = ∠RMS     [Since, angles opposite to equal sides of a triangle are equal]

∠QPS = ∠RQN = ∠SRM     [Co-interior angle]

∠SMR + ∠SRM = ∠PSR      [Exterior angle property]   -    (1)

∠SRQ + ∠PSR = 180°          [Since, sum of co-interior angles of a parallelogram is equal to 180°]

⇒ ∠SRQ + ∠SMR + ∠SRM = 180°    [From (1)]

⇒ ∠SRQ = 180° - ∠SMR - ∠SRM     -     (2)

∠QRM = ∠SRQ + ∠SRM      -     (3)

∠NRQ + ∠QRM = 180°        [Linear Pair]

⇒∠NRQ + ∠SRQ + ∠SRM = 180°     [From (3)]

⇒∠NRQ + 180° - ∠SMR - ∠SRM + ∠SRM = 180°    [From (2)]

⇒∠NRQ = ∠SMR

Since, two angels of both triangles are equal and  ∠SRM = ∠SMR,

∠NRQ = ∠SMR = ∠SRM = ∠RNQ    

Therefore, line QR = QN     [ Sides opposite to equal angles of a triangles are equal]