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E is the point of intersection of the diagonals and F is the point of intersection of the extensions of sides BC and AD. is a circle whose diameter is the segment, EF. P and Q are Pascal points formed by the circle . Triangles FAB and FCD are similar.
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 four sides can be split into two pairs of adjacent equal-length sides. [7] One diagonal crosses the midpoint of the other diagonal at a right angle, forming its perpendicular bisector. [9] (In the concave case, the line through one of the diagonals bisects the other.) One diagonal is a line of symmetry.
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.
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.
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]
(since these are angles that a transversal makes with parallel lines AB and DC). Also, side AB is equal in length to side DC, since opposite sides of a parallelogram are equal in length. Therefore, triangles ABE and CDE are congruent (ASA postulate, two corresponding angles and the included side). Therefore, =
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]