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 $△ P Q R, P Q + P R > Q R$ $∵$ sum of the two sides is greater than the third side.
$∴ P Q + P R > Q T + T R$ $∵ Q R = Q T + T R$
$\Rightarrow P Q + P R > Q T + T S$ ..... $(1)$ $∵ T R = T S$
In $△ Q S T$ , $Q T + T S > Q S$ ..... $(2)$
$∴$ from $(1)$ and $(2)$ we have
$P Q + P R > Q S$
Hence proved

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