Question

hey guys I please help me!!!

Answer 1

Given,

PR > PQ and 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)]