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Question
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
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Solution
In ΔPQR,
PQ + PR > QR ....(∵ Sum of two sides of a triangle is always greater than the third side.) ....(i)
Also, in ΔSQR,
SQ + SR > QR ....(∵ Sum of two sides of a triangle is always greater than the third side.) ....(ii)
Dividing (i) by (ii),
`"PQ + PR"/"SQ + SR" > "QR"/"QR"`
`"PQ + PR"/"SQ + SR" > 1`
PQ + PR > SQ + SR
i.e. SQ + SR < PQ + PR.
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