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The interior perpendicular bisector of a side of a triangle is the segment, falling entirely on and inside the triangle, of the line that perpendicularly bisects that side. The three perpendicular bisectors of a triangle's three sides intersect at the circumcenter (the center of the circle through the three vertices). Thus any line through a ...
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.
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.
The set of points equidistant from two points is a perpendicular bisector to the line segment connecting the two points. [8] The set of points equidistant from two intersecting lines is the union of their two angle bisectors. All conic sections are loci: [9] Circle: the set of points at constant distance (the radius) from a fixed point (the ...
Perpendicular bisector of a line segment. The point where the red line crosses the black line segment is equidistant from the two end points of the black line segment. The cyclic polygon P is circumscribed by the circle C. The circumcentre O is equidistant to each point on the circle, and a fortiori to each vertex of the polygon.
The perpendicular bisector of a chord passes through the centre of a circle; equivalent statements stemming from the uniqueness of the perpendicular bisector are: A perpendicular line from the centre of a circle bisects the chord. The line segment through the centre bisecting a chord is perpendicular to the chord.
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 ...
In general the centre O of a circle on which points P and Q lie must be such that OP and OQ are equal distances. Therefore O must lie on the perpendicular bisector of the line segment PQ. [1] For n distinct points there are n(n − 1)/2 bisectors, and the concyclic condition is that they all meet in a single point, the centre O.