Is it possible to say if the intersection of perpendicular bisectors of sides of a triangle is outside it,it has an angle more than $90$?
I have saw the lemma that is reverse of it But I can't find a proof for that too.This is a lemma that if it is true my question will solved using this if you want I can put the whole question.
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$\begingroup$A comparative/comprehensive situation helps as the biggest angle $C$ lies opposite the largest side $c$, we consider inequalities comparison with right triangle at center of a set of three.
Draw three triangles $ CAB $ in a circle with a common vertex at $C$ making obtuse, right and acute angles and three variable sides $c = AB.$
Perpendicular bisectors of sides in each case meet at circle center/circumcenter $O$ which in these three cases lie outside the triangle, on the hypotenuse and inside the triangle respectively.
$\endgroup$ $\begingroup$The intersection of the perpendicular bisectors of a triangle is the circumcenter. If the triangle has all angles acute, the circumcenter is within the triangle, and conversely. If the triangle is right, the circumcenter is on the hypotenuse (since the hypotenuse is the circumcenter's diameter), and conversely. Therefore, if the triangle is obtuse, the point lies outside the triangle, and conversely.
$\endgroup$ $\begingroup$Consider the intersection of the perpendicular bisectors of the longest side and either of the shorter sides. Note that joining the midpoint of these two sides makes a smaller triangle similar to the full triangle.
Now for an obtuse triangle, the perpendicular bisector of the shorter side will start inside this midpoint triangle and hence will cross the longest side before its midpoint, putting the intersection of those two bisectors outside the triangle.
(The following is not a formal proof, though it could be developed into one with some work.)
Thought experiment (building upon @JohnWaylandBales' comment): imagine a triangle $\triangle ABC$ with $B,C$ fixed and $A$ a freely moving point. Let $O$ be the circumcenter, then $O$ can lie on the line $BC$ iff $OB=OC$ by definition of the circumcenter i.e. $O$ must be midpoint of segment $BC\,$. In that case $\widehat{A} = 90^\circ$ since it subtends diameter $BC\,$. It is intuitively obvious that $O$ depends continuously on $A$, so the angle $\widehat A$ only ever changes from $\lt 90^\circ$ to $\gt 90^\circ$ when $O$ crosses the line $BC$. It follows that $O$ is on the same side of $BC$ with $A$ iff $\widehat A \lt 90^\circ$. Finally, $O$ is inside triangle $\triangle ABC$ iff it is on the same side with any vertex relative to the opposite side which, from the above, is equivalent to all angles being acute. Conversely, it is outside iff at least one angle is obtuse.
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