Question

S is any point on side QR of a $\Delta PQR$. Show that : PQ + QR + RP > 2 PS

Answer 1

Theorem: In a triangle , sum of the length of any two sides is greater than the third side.

Draw the triangle and join the points $P$ and $S$

In the figure,

In $\Delta PQS$, according to the theorem,

$PQ+QS>PS$ ........... (1)

In $\Delta PSR$, according to the theorem,

$PR+SR>PS$ ........... (2)

Adding (1) and (2),

$PQ+QS+SR+PR>PS+PS$

$PQ+(QS+SR)+PR>2PS$

$PQ+QR+PR>2PS$