Question

is the perpendicular bisector of each side concurrent ​

Answer 1

Answer:

Let's prove that the three perpendicular bisectors of the sides of a triangle are concurrent which means that they intersect at one point. ... If a line is a perpendicular bisector of the side of a triangle, then it bisects the side into two halves and forms right angles with the side.

Answer 2

Answer:

Let’s prove that the three perpendicular bisectors of the sides of a triangle are concurrent which means that they intersect at one point.

To do so, let’s consider triangle ABC as constructed in Figure 1. In triangle ABC, perpendicular bisectors FD and FE have been constructed. If a line is a perpendicular bisector of the side of a triangle, then it bisects the side into two halves and forms right angles with the side. Therefore, the following can be determined from the figure…

AD = DC and AE = EB

Angles FDA, FDC, FEA and FEB are each 90 degrees.perpendicular bisectors FD and FE intersect at point F. To prove that the three perpendicular bisectors of triangle ABC are concurrent, we must show that the third perpendicular bisector goes through point F as well.

For purposes of convenience, perpendicular bisectors DF and FE have been shortened to segments FD and FEcan be concluded then that all three perpendicular bisectors, FD, FE, and FG, are concurrent at point F because point F is equidistant from all three vertices of the triangle. This point is also called the circumcenter because it is the center of the circle that circumscribes the triangle.Since FC = FB, this means that point F must be equidistant from points C and B as well. Since points C and B are the endpoints of segment BC and point F is equidistant from those points, it can be concluded that point F lies on the perpendicular bisector of side BC. That perpendicular bisector has been constructed as segment FGthe radii of the circle are FA, FB, and FC.