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Question

In fig(iv), T is a point on side QR of triangle PQR and S is a point such that RT = ST. Prove that PQ + PR > QS
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Solution

In $$\triangle{PQR}, PQ+PR>QR$$ $$\because$$ sum of the two sides is greater than the third side.
$$\therefore PQ+PR>QT+TR$$   $$\because QR=QT+TR$$
$$\Rightarrow PQ+PR>QT+TS$$  .....$$\left(1\right)$$   $$\because TR=TS$$
In $$\triangle QST$$,$$QT+TS>QS$$      .....$$\left(2\right)$$
$$\therefore$$ from $$\left(1\right)$$ and $$\left(2\right)$$ we have
$$PQ+PR>QS$$
Hence proved 

Mathematics

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