Question

please give me the answer only if you know..

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Answer 1

Answer:

$\huge\red{\boxed{\sf Answer}}$

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°

⇒ 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 △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 ]

$\huge\orange{\boxed{\sf Hence\: Proved}}$