T is a point on QR of triangle PQR.S is a point such that ST=TR.prove PQ+PR>QS

Question

tell which congruent is it.

Answer 1

?????????????? PQR.S ???????????????

Answer 2

To solve this problem, we use the triangle inequality that states any side of a triangle is always shorter than the sum of the other two sides.   So that, in triangle PQR, QR<PQ+PR ··· Ex.1

Draw a circle centered at T with radius TR, and mark a point S at anywhere on the circle, so ST=TR ··· Ex.2

Cond.1 where points S,Q and T form a triangle SQT.   From the triangle inequality, QS<QT+ST.   From Ex.2, QT+ST=QT+TR=QR, so that QS<QR.   Thus, with Ex.1, we have: QS<PQ+PR

Cond.2 where point S is anywhere on a ray drawn from T thru Q, so QS=|ST-QT|.  From Ex.2, |ST-QT|=|TR-QT|, and |TR-QT|<QR, so that QS<QR.   Thus, with Ex.1, we have: QS<PQ+PR

Therefore, from cond.1 and cond.2, we have: PQ+PR>QS