Question

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

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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