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The circle meets the angle at two points: one on each leg. Using each of these points as a center, draw two circles of the same size. The intersection of the circles (two points) determines a line that is the angle bisector. The proof of the correctness of this construction is fairly intuitive, relying on the symmetry of the problem.
The value of the two products in the chord theorem depends only on the distance of the intersection point S from the circle's center and is called the absolute value of the power of S; more precisely, it can be stated that: | | | | = | | | | = where r is the radius of the circle, and d is the distance between the center of the circle and the ...
The intersection points of this circle with the two given lines (5) are T1 and T2. Two circles of the same radius, centered on T1 and T2, intersect at points P and Q. The line through P and Q (1) is an angle bisector. Rays have one angle bisector; lines have two, perpendicular to one another.
Next to the intersecting chords theorem and the tangent-secant theorem, the intersecting secants theorem represents one of the three basic cases of a more general theorem about two intersecting lines and a circle - the power of point theorem.
Creating the one point or two points in the intersection of two circles (if they intersect). For example, starting with just two distinct points, we can create a line or either of two circles (in turn, using each point as centre and passing through the other point). If we draw both circles, two new points are created at their intersections.
Viète used this approach to shrink one of the given circles to a point, thus reducing the problem to a simpler, already solved case. He first solved the CLL case (a circle and two lines) by shrinking the circle into a point, rendering it an LLP case. He then solved the CLP case (a circle, a line and a point) using three lemmas.
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.
For three non-collinear points, these two lines cannot be parallel, and the circumcenter is the point where they cross. Any point on the bisector is equidistant from the two points that it bisects, from which it follows that this point, on both bisectors, is equidistant from all three triangle vertices.