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Question

In figure, $$T$$ is a point on side $$QR$$ of $$\Delta PQR$$ and $$S$$ is a point such that $$RT = ST$$. Prove that $$PQ + PR > QS$$.
1878911_24d48170d3ce40caa073b10dbc6aea6a.png


Solution

In $$\Delta PQR$$, we have
$$PQ + PR > QR$$  (the sum of two sides of a triangle is greater than third side).
$$PQ + PR > QT + TR$$   $$[\because QT + TR = QR]$$
$$PQ + PR > QT + ST$$  …(i)    $$[\because TR = ST]$$
In $$\Delta QST$$, we have
$$QT + ST > QS$$  …(ii)
From (i) and (ii), we get
$$PQ + PR > QS$$.
Hence proved.

Mathematics

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