Not always because the diagonals of a rectangle bisect each other but they are not perpendicular to each other.
any isosceles triangle
Yes, in a circle, the perpendicular bisector of a chord does indeed pass through the center of the circle. This is because the perpendicular bisector of a chord divides it into two equal segments and is equidistant from the endpoints of the chord. Since the center of the circle is the point that is equidistant from all points on the circle, it must lie on the perpendicular bisector. Thus, any chord's perpendicular bisector will always intersect the center of the circle.
With a straight-edge and a compass:Swing arcs from each end of the segment with the compass (without changing the settings)Connect the intersections of these arcs.The resultant is a perpendicular bisector of the segment.
Can't be sure what you're asking, but it would be two circles with equal radii longer that half the length of the segment, using the endpoints as the origins of the two circles.
If line BE is the bisector of segment AC, it means that BE divides AC into two equal segments. Therefore, if AB is 7, then AC must be twice that length, making AC equal to 14.
So that the arc is mid-way in perpendicular to the line segment
Equilateral triangles
iscoceles triangle! =)
any isosceles triangle
It bisects the line segment at midpoint at 90 degrees and its slope is the reciprocal of the line segment's slope plus or minus.
Yes, in a circle, the perpendicular bisector of a chord does indeed pass through the center of the circle. This is because the perpendicular bisector of a chord divides it into two equal segments and is equidistant from the endpoints of the chord. Since the center of the circle is the point that is equidistant from all points on the circle, it must lie on the perpendicular bisector. Thus, any chord's perpendicular bisector will always intersect the center of the circle.
With a straight-edge and a compass:Swing arcs from each end of the segment with the compass (without changing the settings)Connect the intersections of these arcs.The resultant is a perpendicular bisector of the segment.
The perpendicular bisector of the line XY will meet it at its midpoint at right angles.
Can't be sure what you're asking, but it would be two circles with equal radii longer that half the length of the segment, using the endpoints as the origins of the two circles.
The answer letters always rearrange so here are the answers point H is the midpoint of FG line t intersects FG at a right angle Line T is perpendicular to FG
If line BE is the bisector of segment AC, it means that BE divides AC into two equal segments. Therefore, if AB is 7, then AC must be twice that length, making AC equal to 14.
No. Since a line is infinite, it has no mid-point. A bisector must go through a midpoint so nothing can bisect a line (not even a segment).