Question

in the give figure, ∠PQR= ∠PRQ
prove that, ∠PSQ=PRT


NOTE- solve it step by step ✨​

Answer 1

||GIVEN||

• ∆PQR, is standing on the base line ST.
• ∠PQR = ∠PRQ.

||TO PROVE||

• ∠PQS = ∠PRT.

||REQUIRED PROOF||

∠PQR+∠PQS = 180° [Linear Pair]

∠PQS = 180°-∠PQR → (1)

∠PRQ+∠PRT = 180° [Linear Pair]

∠PRT = 180°-∠PRQ → (2)

By Equating (1) & (2)

As, ∠PQR = ∠PRQ [Given]

180°-∠PQR = 180°-∠PRQ

180°-∠PQR = 180°-∠PQR

∠PQS = ∠PRT

• $\longmapsto\;\bold{∠PQS = ∠PRT}$

Hence, Proved.

$\tt{@MrMonarque}$

Hope It Helps You ✌️

Answer 2

GIVEN :-

☞ ∆PQR, is standing on the base line ST.

☞ ∠PQR = ∠PRQ.

TO PROVE :-

$\sf ∠PSQ = ∠PRT$

PROOF :-

$\sf ∠PQR+∠PQS = 180° [Linear \:Pair]$

$\sf ∠PQS = 180°-∠PQR ⇒ (1)$

$\sf ∠PRQ+∠PRT = 180° [Linear Pair]$

$\sf ∠PRT = 180°-∠PRQ ⇒ (2)$

By Equating (1) & (2)

$\sf As, ∠PQR = ∠PRQ [Given]$

$\sf 180°-∠PQR = 180° - ∠PRQ$

$\sf 180°-∠PQR = 180° - ∠PQR$

$\sf 180°-180°-∠PQR = - ∠PQR$

$\sf -∠PQR = - ∠PQR$

$\sf ∠PQS = ∠PRQ$

∴ ∠PQS = ∠PRQ

∴ $\fbox {\sf∠PSQ = ∠PRT \: [CPCT]}$

Hence Proved.

$\fbox \orange{ \sf @IIMrVelvetII}$

$\fbox \green{ \sf \: Thank You!!!}$