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For a triangle ABC, let l an arbitrary line and A 0 B 0 C 0 the Cevian triangle of an arbitrary point P. l intersects BC, CA, and AB at A 1, B 1, and C 1 respectively. Then AA 1 ∩B 0 C 0, BB 1 ∩C 0 A 0, and CC 1 ∩A 0 B 0 are colinear. If P is the centroid of the triangle ABC, the line is Newton-Gauss line of the quadrilateral composed of ...
Given two points of interest, finding the midpoint of the line segment they determine can be accomplished by a compass and straightedge construction.The midpoint of a line segment, embedded in a plane, can be located by first constructing a lens using circular arcs of equal (and large enough) radii centered at the two endpoints, then connecting the cusps of the lens (the two points where the ...
Hyperbola: the midpoints of parallel chords lie on a line. Hyperbola: the midpoint of a chord is the midpoint of the corresponding chord of the asymptotes. The midpoints of parallel chords of a hyperbola lie on a line through the center (see diagram). The points of any chord may lie on different branches of the hyperbola.
In geometry, symmedians are three particular lines associated with every triangle.They are constructed by taking a median of the triangle (a line connecting a vertex with the midpoint of the opposite side), and reflecting the line over the corresponding angle bisector (the line through the same vertex that divides the angle there in half).
The line perpendicular to the directrix and passing through the focus (that is, the line that splits the parabola through the middle) is called the "axis of symmetry". The point where the parabola intersects its axis of symmetry is called the "vertex" and is the point where the parabola is most sharply curved. The distance between the vertex ...
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In projective geometry, the harmonic conjugate point of a point on the real projective line with respect to two other points is defined by the following construction: Given three collinear points A, B, C, let L be a point not lying on their join and let any line through C meet LA, LB at M, N respectively.
The two circles in the Two points, one line problem where the line through P and Q is not parallel to the given line l, can be constructed with compass and straightedge by: Draw the line m through the given points P and Q. The point G is where the lines l and m intersect; Draw circle C that has PQ as diameter. Draw one of the tangents from G to ...