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A quadrilateral such as BCEF is called an adventitious quadrangle when the angles between its diagonals and sides are all rational angles, angles that give rational numbers when measured in degrees or other units for which the whole circle is a rational number. Numerous adventitious quadrangles beyond the one appearing in Langley's puzzle have ...
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.
Diagonal intersection is a term used in mathematics, especially in set theory. If δ {\displaystyle \displaystyle \delta } is an ordinal number and X α ∣ α < δ {\displaystyle \displaystyle \langle X_{\alpha }\mid \alpha <\delta \rangle } is a sequence of subsets of δ {\displaystyle \displaystyle \delta } , then the diagonal intersection ...
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.
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]
Denote the segments that the diagonal intersection P divides diagonal AC into as AP = p 1 and PC = p 2, and similarly P divides diagonal BD into segments BP = q 1 and PD = q 2. Then the quadrilateral is tangential if and only if any one of the following equalities are true: [30]
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)