5. In Fig 7.51, PR> PQ and PS bisects 2 QPR. ProvethatPSRäťĽPSQ.

1.	5. In Fig 7.51, PR> PQ and PS bisects 2 QPR. ProvethatPSRäťĽPSQ.
Answer» 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)]

Answer»

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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