I assume that's what the OP means by the method of perpendicular bisectors but it seems like the easiest way to do it.
32.
Three of them are the concurrent at the triangle's centroid; indeed, they are the only area bisectors that go through the centroid.
33.
The three perpendicular bisectors of a triangle's three sides intersect at the circumcenter ( the center of the circle through the three vertices ).
34.
Construct a circle through points A, B and C by finding the perpendicular bisectors ( red ) of the sides of the triangle ( blue ).
35.
If the internal bisectors of angles A, B, and C have lengths t _ a, t _ b, and t _ c, then
36.
Vi�te began by solving the "'PPP "'case ( three points ) following the method of Euclid in his " angle bisectors.
37.
The internal angle bisectors are segments in the interior of the triangle reaching from one vertex to the opposite side and bisecting the vertex angle into two equal angles.
38.
The Spieker center is also the point where all three cleavers of the triangle ( perimeter bisectors with an endpoint at a side's midpoint ) intersect each other.
39.
The center of the incircle, called the "'incenter "', can be found as the intersection of the three external bisectors of the other two.
40.
Since there are two such bisectors at every intersection point of the three given lines, there are four solutions to the general "'LLL "'problem.
