Question

In adjacent figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

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)]

Hope this will help you.......

Answer 2

Hope it will help you please mark it as brainliest