Question

# Question 5 In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ .

Solution

## PR>PQ                           (Given) ∠PQR > ∠PRQ.........(1) (In any triangle, the angle opposite to the longer side is larger.) We also have ∠PQR+∠QPS+∠PSQ=180∘ (Angle sum property of triangle) ∴∠PQR=180∘−∠QPS−∠PSQ.....(2) And, ∠PRQ+∠RPS+∠PSR=180∘   (Angle sum property of triangle) ∴∠PRQ=180∘−∠PSR−∠RPS .........(3) Substituting (2) and (3) in (1), we get 180∘−∠QPS−∠PSQ > 180∘−∠PSR−∠RPS We also have ∠QPS=∠RPS, using this in the above equation, we get −∠PSQ > −∠PSR ⇒∠PSQ < ∠PSR

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