In Delta PQR,S is any point in its interior. | vedclass

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(N/A) $1$. Extend $QS$ to meet $PR$ at point $T$.$2$. In $\Delta PQT$,by the triangle inequality theorem,the sum of two sides is greater than the thir

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(N/A) $1$. Extend $QS$ to meet $PR$ at point $T$.
$2$. In $\Delta PQT$,by the triangle inequality theorem,the sum of two sides is greater than the third side: $PQ + PT > QT$. This can be written as $PQ + PT > QS + ST$ (Equation $1$).
$3$. In $\Delta SRT$,by the triangle inequality theorem,$ST + TR > SR$ (Equation $2$).
$4$. Adding Equation $1$ and Equation $2$: $PQ + PT + ST + TR > QS + ST + SR$.
$5$. Simplifying by canceling $ST$ from both sides: $PQ + (PT + TR) > QS + SR$.
$6$. Since $PT + TR = PR$,we get $PQ + PR > QS + SR$,which is equivalent to $SQ + SR < PQ + PR$.

In $\Delta PQR$,$S$ is any point in its interior. Prove that $SQ + SR < PQ + PR$.

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