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In geometry, the perpendicular bisector construction of a quadrilateral is a construction which produces a new quadrilateral from a given quadrilateral using the perpendicular bisectors to the sides of the former quadrilateral.
The line determined by the points of intersection of the two circles is the perpendicular bisector of the segment. Because the construction of the bisector is done without the knowledge of the segment's midpoint , the construction is used for determining as the intersection of the bisector and the line segment.
Constructing the perpendicular bisector from a segment; Finding the midpoint of a segment. Drawing a perpendicular line from a point to a line. Bisecting an angle; Mirroring a point in a line; Constructing a line through a point tangent to a circle; Constructing a circle through 3 noncollinear points; Drawing a line through a given point ...
For one other site , the points that are closer to than to , or equally distant, form a closed half-space, whose boundary is the perpendicular bisector of line segment . Cell R k {\displaystyle R_{k}} is the intersection of all of these n − 1 {\displaystyle n-1} half-spaces, and hence it is a convex polygon . [ 6 ]
The perpendicular bisectors of all chords of a circle are concurrent at the center of the circle. The lines perpendicular to the tangents to a circle at the points of tangency are concurrent at the center. All area bisectors and perimeter bisectors of a circle are diameters, and they are concurrent at the circle's center.
To construct the perpendicular bisector of the line segment between two points requires two circles, each centered on an endpoint and passing through the other endpoint (operation 2). The intersection points of these two circles (operation 4) are equidistant from the endpoints. The line through them (operation 1) is the perpendicular bisector.
Carnot's theorem: if three perpendiculars on triangle sides intersect in a common point F, then blue area = red area. Carnot's theorem (named after Lazare Carnot) describes a necessary and sufficient condition for three lines that are perpendicular to the (extended) sides of a triangle having a common point of intersection.
More specifically, let A, B, C and D be four points on a circle such that the lines AC and BD are perpendicular. Denote the intersection of AC and BD by M. Drop the perpendicular from M to the line BC, calling the intersection E. Let F be the intersection of the line EM and the edge AD. Then, the theorem states that F is the midpoint AD.