The perpendicular bisector of the line XY will meet it at its midpoint at right angles.
Biconditional Statement for: Perpendicular Bisector Theorem: A point is equidistant if and only if the point is on the perpendicular bisector of a segment. Converse of the Perpendicular Bisector Theorem: A point is on the perpendicular bisector of the segment if and only if the point is equidistant from the endpoints of a segment.
I believe the answer is "perpendicular line". Forgive me if I'm wrong :)
BC and DE
The arcs must intersect because you need a point to use with the point of the angle's vertex to make the line that intersects the angle.
Drawing perpendicular bisector for a line:Place the sharp end of a pair of compasses at one end of the line, and open it to just over half of the line. Draw an arc which must intersect the line in the position described. Then put the sharp end at the other of the line and, keeping the compassing at the same length, draw another arc which intersects the first one twice and also the line. Then draw a straight line through the two places where the arcs intersect. This line is the perpendicular bisector. Drawing perpendicular bisector of angle:Places the sharp end of the compass at the point of the angle and, after having opened it arbitraily wide, draw an arc which intersects the two lines meeting to form the angle each once in the said position. Then remove the compass and, always keeping it opened at the SAME length, place the sharp end at each of the two places where the previous arc cuts each of the two lines meeting to form the angle. In this position with the described length, draw a small arc at each of the said places, which should cross each other. Draw a straight line from the point of the angle to this crossing. This should be the bisector of the angle.
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
Biconditional Statement for: Perpendicular Bisector Theorem: A point is equidistant if and only if the point is on the perpendicular bisector of a segment. Converse of the Perpendicular Bisector Theorem: A point is on the perpendicular bisector of the segment if and only if the point is equidistant from the endpoints of a segment.
Converse of the Perpendicular Bisector Theorem - if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.Example: If DA = DB, then point D lies on the perpendicular bisector of line segment AB.you :))
on the perpendicular bisector
If a point is on the perpendicular bisector of a segment, then it is equidistant, or the same distance, from the endpoints of the segment.
on the perpendicular bisector of the segment.
The converse of perpendicular bisector theorem states that if a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
The locus point is the perpendicular bisector of AB. The locus point is the perpendicular bisector of AB.
Not enough information has been given but if a given line bisects another line at 90 degrees then it is the perpendicular bisector of that line.
Adjust a compass so the distance between the point and the pencil is more than half of the length of the segment. With the point at one end of the segment draw an arc that intersects the segment. Without adjusting the compass, with the point at the other end of the segment draw an arc that intersects the first arc at two places. The line that includes those two intersecting points is the perpendicular bisector.
I believe the answer is "perpendicular line". Forgive me if I'm wrong :)
BC and DE