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Not always because the diagonals of a rectangle bisect each other but they are not perpendicular to each other.

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10y ago

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When constructing a perpendicular bisector why must the compass opening be greater than the length of the segment?

So that the arc is mid-way in perpendicular to the line segment


To construct a perpendicular bisector to a given line segment one must construct two?

Equilateral triangles


Which types of triangles must always have at least one angle bisector that is also a perpendicular bisector?

iscoceles triangle! =)


What triangle must always have at least one angle bisector that is also a perpendicular bisector?

any isosceles triangle


What must be true about a perpendicular bisector and the segment it bisects?

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.


What must you do to construct the midpoint of a segment?

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.


How is line m is the perpendicular bisector of XY. If line m intersects XY at point Z statements must be true?

The perpendicular bisector of the line XY will meet it at its midpoint at right angles.


To construct a perpendicular bisector to a given line segment one must construct two what?

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.


What Line t is the perpendicular bisector of FG. If line t intersects FG at point H statements must be true Check all that apply.?

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


Line BE is the bisector of segment AC. If AB 7 then AC .?

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.


Can a segment bisect a line?

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).


How do you construct a perpendicular bisector of each side of a parallelogram?

With a compass and a straight edge and the lines must bisect each other at 90 degrees