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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 ...
Circles tangent to two given points must lie on the perpendicular bisector. Circles tangent to two given lines must lie on the angle bisector. Tangent line to a circle from a given point draw semicircle centered on the midpoint between the center of the circle and the given point. Power of a point and the harmonic mean [clarification needed]
Newton formulates Apollonius' problem as a problem in trilateration: to locate a point Z from three given points A, B and C, such that the differences in distances from Z to the three given points have known values. [31] These four points correspond to the center of the solution circle (Z) and the centers of the three given circles (A, B and C).
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 ...
The ratio x : y is the ratio of the perpendicular distances from the point to the sides (extended if necessary) opposite vertices A and B respectively; the ratio y : z is the ratio of the perpendicular distances from the point to the sidelines opposite vertices B and C respectively; and likewise for z : x and vertices C and A.
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.
For any triangle, and, in particular, any right triangle, there is exactly one circle containing all three vertices of the triangle. (Sketch of proof. 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.
As mentioned above, every triangle has a unique circumcircle, a circle passing through all three vertices, whose center is the intersection of the perpendicular bisectors of the triangle's sides. Furthermore, every triangle has a unique Steiner circumellipse, which passes through the triangle's vertices and has its center at the triangle's ...