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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 ...
Consider a triangle ABC.Let the angle bisector of angle ∠ A intersect side BC at a point D between B and C.The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment CD is equal to the ratio of the length of side AB to the length of side AC:
The perpendicular bisector of a side of a triangle is the line that is perpendicular to that side and passes through its midpoint. The three perpendicular bisectors of a triangle's three sides intersect at the circumcenter (the center of the circle through the three vertices).
The locus of points equidistant from two given points is a straight line that is called the perpendicular bisector of the line segment connecting the points. The perpendicular bisectors of any two sides of a triangle intersect in exactly one point. This point must be equidistant from the vertices of the triangle.)
the perpendicular bisectors p a, p b, and p c of the sides (each being the length of a segment perpendicular to one side at its midpoint and reaching to one of the other sides); the lengths of line segments with an endpoint at an arbitrary point P in the plane (for example, the length of the segment from P to vertex A is denoted PA or AP);
The three perpendicular bisectors meet at the circumcenter. Other sets of lines associated with a triangle are concurrent as well. For example: Any median (which is necessarily a bisector of the triangle's area) is concurrent with two other area bisectors each of which is parallel to a side. [1]
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.
To make the perpendicular to the line AB through the point P using compass-and-straightedge construction, proceed as follows (see figure left): Step 1 (red): construct a circle with center at P to create points A' and B' on the line AB, which are equidistant from P. Step 2 (green): construct circles centered at A' and B' having equal radius.