Question

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

Answer 1

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.