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The bisectors of the angles at B and D intersect on the diagonal AC. A diagonal BD of the quadrilateral is a symmedian of the angles at B and D in the triangles ∆ ABC and ∆ ADC. The point of intersection of the diagonals is located towards the sides of the quadrilateral to proportional distances to the length of these sides.
where P is the point of intersection of the diagonals. From this equation it follows almost immediately that the diagonals of a convex quadrilateral are perpendicular if and only if the projections of the diagonal intersection onto the sides of the quadrilateral are the vertices of a cyclic quadrilateral. [6]
In geometry, Brahmagupta's theorem states that if a cyclic quadrilateral is orthodiagonal (that is, has perpendicular diagonals), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side. [1] It is named after the Indian mathematician Brahmagupta (598-668). [2]
One more interesting line (in some sense dual to the Newton's one) is the line connecting the point of intersection of diagonals with the vertex centroid. The line is remarkable by the fact that it contains the (area) centroid. The vertex centroid divides the segment connecting the intersection of diagonals and the (area) centroid in the ratio 3:1.
If the incircle is tangent to the sides AB, BC, CD, DA at T 1, T 2, T 3, T 4 respectively, and if N 1, N 2, N 3, N 4 are the isotomic conjugates of these points with respect to the corresponding sides (that is, AT 1 = BN 1 and so on), then the Nagel point of the tangential quadrilateral is defined as the intersection of the lines N 1 N 3 and N ...
More generally, if the quadrilateral is a rectangle with sides a and b and diagonal d then Ptolemy's theorem reduces to the Pythagorean theorem. In this case the center of the circle coincides with the point of intersection of the diagonals. The product of the diagonals is then d 2, the right hand side of Ptolemy's relation is the sum a 2 + b 2.
For a cyclic quadrilateral that is also orthodiagonal (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengths p 1 and p 2 and divides the other diagonal into segments of lengths q 1 and q 2. Then [28] (the first equality is Proposition 11 in Archimedes' Book of Lemmas)
If M, N are the midpoints of the diagonals, and E, F are the intersection points of the extensions of opposite sides, then the area can also be expressed as = ¯ ¯ ¯ where Q is the foot of the perpendicular to the line EF through the center of the incircle. [9]