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

1
by nave
meaning of SQ &It: PQ

2014-08-12T00:42:10+05:30
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.
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