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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.

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Q: How do you draw a perpendicular bisector?
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Related questions

How do you draw a perpendicular bisector of an angle?

The right way


How do you draw parallel perpendicular bisector?

Two lines cannot be parallel and perpendicular at the same time.


The first step in the construction of a perpendicular bisector is to draw a?

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An angle bisector bisects an angle. A perpendicular bisector bisects a side.


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haterz gonna hate but it is yes


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State the Perpendicular Bisector Theorem and its converse as a biconditional?

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.


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The Perpendicular bisector concurrency conjecture is the circumcenter


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