Question

Pq > pr , how to prove angle psr = 1/2(angle pqr+ angle prq)

Answer 1

Given:
PR > PQ & PS bisects ∠QPR
To prove:
∠PSR > ∠PSQ
Proof:
∠PQR > ∠PRQ — (i) (PR > PQ as angle opposite to larger side is larger.)
∠QPS = ∠RPS — (ii) (PS bisects ∠QPR)
∠PSR = ∠PQR +∠QPS — (iii)
(exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS — (iv)
(exterior angle of a triangle equals to the sum of opposite interior angles)
Adding (i) and (ii)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]
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Hope this will help you.......

Answer 2

Question :-

In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer :-

ST is a straight line.

∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair]

Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair]

From (1) and (2), we have

∠PQS + ∠PQR = ∠PRT + ∠PRQ

But ∠PQR = ∠PRQ [Given]

∴ ∠PQS = ∠PRT

Plz mrk as brainliest ❤