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

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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