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Euler's quadrilateral theorem or Euler's law on quadrilaterals, named after Leonhard Euler (1707–1783), describes a relation between the sides of a convex quadrilateral and its diagonals. It is a generalisation of the parallelogram law which in turn can be seen as generalisation of the Pythagorean theorem .
A self-intersecting quadrilateral is called variously a cross-quadrilateral, crossed quadrilateral, butterfly quadrilateral or bow-tie quadrilateral. In a crossed quadrilateral, the four "interior" angles on either side of the crossing (two acute and two reflex , all on the left or all on the right as the figure is traced out) add up to 720°.
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]
Usually the quadrilateral is assumed to be convex, but there are also crossed cyclic quadrilaterals. The formulas and properties given below are valid in the convex case. The word cyclic is from the Ancient Greek κύκλος (kuklos), which means "circle" or "wheel". All triangles have a circumcircle, but not all quadrilaterals do.
This formula generalizes Heron's formula for the area of a triangle. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as d (or any one side) approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
Bretschneider's formula generalizes Brahmagupta's formula for the area of a cyclic quadrilateral, which in turn generalizes Heron's formula for the area of a triangle.. The trigonometric adjustment in Bretschneider's formula for non-cyclicality of the quadrilateral can be rewritten non-trigonometrically in terms of the sides and the diagonals e and f to give [2] [3]
But two of these diagonals are the same as the diagonals of the tangential quadrilateral, and the third diagonal of the hexagon is the line through two opposite points of tangency. Repeating this same argument with the other two points of tangency completes the proof of the result.
An arbitrary quadrilateral and its diagonals. Bases of similar triangles are parallel to the blue diagonal. Ditto for the red diagonal. The base pairs form a parallelogram with half the area of the quadrilateral, A q, as the sum of the areas of the four large triangles, A l is 2 A q (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, A s is a quarter of A ...