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Ptolemy's Theorem yields as a corollary a pretty theorem [2] regarding an equilateral triangle inscribed in a circle. Given An equilateral triangle inscribed on a circle, and a point on the circle. The distance from the point to the most distant vertex of the triangle is the sum of the distances from the point to the two nearer vertices.
Solution of triangles (Latin: solutio triangulorum) is the main trigonometric problem of finding the characteristics of a triangle (angles and lengths of sides), when some of these are known. The triangle can be located on a plane or on a sphere. Applications requiring triangle solutions include geodesy, astronomy, construction, and navigation.
The triangle inequality states that the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side. [48] Conversely, some triangle with three given positive side lengths exists if and only if those side lengths satisfy the triangle inequality. [49]
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 approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
Furthermore, let E and F the midpoints of its diagonals AC and BD and P be the center of its incircle. Given such a configuration the point P is located on the Newton line, that is line EF connecting the midpoints of the diagonals. [1] A tangential quadrilateral with two pairs of parallel sides is a rhombus.
The area (A) of a regular heptagon of side length a is given by: A = 7 4 a 2 cot π 7 ≃ 3.634 a 2 . {\displaystyle A={\frac {7}{4}}a^{2}\cot {\frac {\pi }{7}}\simeq 3.634a^{2}.} This can be seen by subdividing the unit-sided heptagon into seven triangular "pie slices" with vertices at the center and at the heptagon's vertices, and then ...
where ∆(APB) is the area of triangle APB. 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]
The diagonals divide the polygon into 1, 4, 11, 24, ... pieces. [ a ] For a regular n -gon inscribed in a circle of radius 1 {\displaystyle 1} , the product of the distances from a given vertex to all other vertices (including adjacent vertices and vertices connected by a diagonal) equals n .