Question

s is any point in the interior of triangle PQR then show that SQ<PQ  and  PR>SR

Answer 1

Lets say that the angle between PQ and SQ is Ф. Now lets say that foot of perpendicular from S on PQ is T.
Now QT = PQ holds, if and only if, T lies on P and this is possible if and only if, when Ф = 0, but S is interior point so this is not possible.
Thus QT < PQ.
Also QT = QS cos(Ф)
So, QS cos(Ф) < PQ
Now if Ф ∈ (0, 90) then cos(Ф) ∈ (1, 0) Thus,
⇒ QS < ( PQ / cos(Ф) )
And if you remember the fact that when we divide any number by positive number smaller than one, the result is always greater than the number
⇒  ( PQ / cos(Ф) ) > PQ
Thus QS < PQ

And when Ф ∈ (90, 180) then cos(Ф) is negative, so  QS cos(Ф) is negative thus QS < PQ.

Similarly prove the second one.

Answer 2

Thank you for asking this question:

Here is your answer:

First of all we will draw QS so that it can intersect PR at T

From ΔPQT we will have this:

PQ + PT > QT  (this is the sum of any two sides which are comparatively greater than third side)

PQ + PT > SQ + ST

From Δ TSR we will have this:

ST + TR > SR

Now we will add both the equations

PQ + PT  + ST + TR > SQ + ST + SR

PQ + PT + TR > SQ + SR

PQ + PR > SQ + SR

If there is any confusion please leave a comment below.