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In ΔPQR, PR > PQ [∵ Given] Hence, ∠Q > ∠R [∵ In a triangle, longer, side has greater angle opposite to it] Adding ∠RPS both sides, ∠Q + ∠RPS > ∠R + ∠RPS ⇒ ∠Q + ∠RPS > ∠PSQ [∵ ∠PSQ is exterior angle of triangle PSR] ⇒∠Q + ∠QPS > ∠PSQ [∵∠RPS = ∠QPS] ⇒∠PSR > ∠PSQ [∵∠PSR is exterior anRead more
In ΔPQR, PR > PQ [∵ Given] Hence, ∠Q > ∠R [∵ In a triangle, longer, side has greater angle opposite to it] Adding ∠RPS both sides, ∠Q + ∠RPS > ∠R + ∠RPS ⇒ ∠Q + ∠RPS > ∠PSQ [∵ ∠PSQ is exterior angle of triangle PSR] ⇒∠Q + ∠QPS > ∠PSQ [∵∠RPS = ∠QPS] ⇒∠PSR > ∠PSQ [∵∠PSR is exterior angle of triangle PQS]
In Figure PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.
In ΔPQR, PR > PQ [∵ Given] Hence, ∠Q > ∠R [∵ In a triangle, longer, side has greater angle opposite to it] Adding ∠RPS both sides, ∠Q + ∠RPS > ∠R + ∠RPS ⇒ ∠Q + ∠RPS > ∠PSQ [∵ ∠PSQ is exterior angle of triangle PSR] ⇒∠Q + ∠QPS > ∠PSQ [∵∠RPS = ∠QPS] ⇒∠PSR > ∠PSQ [∵∠PSR is exterior anRead more
In ΔPQR,
See lessPR > PQ [∵ Given]
Hence, ∠Q > ∠R [∵ In a triangle, longer, side has greater angle opposite to it]
Adding ∠RPS both sides,
∠Q + ∠RPS > ∠R + ∠RPS
⇒ ∠Q + ∠RPS > ∠PSQ [∵ ∠PSQ is exterior angle of triangle PSR]
⇒∠Q + ∠QPS > ∠PSQ [∵∠RPS = ∠QPS]
⇒∠PSR > ∠PSQ [∵∠PSR is exterior angle of triangle PQS]
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.