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For a quadrilateral which is not cyclic, Ptolemy's theorem becomes an inequality: AB×CD+BC×DA>AC×BD. The Ptolemy inequality is still valid when ABCD is a triangular pyramid ...

For a cyclic quadrilateral, the sum of the products of the two pairs of opposite sides equals the product of the diagonals AB×CD+BC×DA=AC×BD (1) (Kimberling 1998, p. 223). ...

Consider two directly similar triangles DeltaA_1B_1C_1 and DeltaA_2B_2C_2 with B_1C_1:A_1C_1:A_1B_1=B_2C_2:A_2C_2:A_2B_2=a:b:c. Then a·A_1A_2, b·B_1B_2 and c·C_1C_2 form the ...

Let the opposite sides of a convex cyclic hexagon be a, a^', b, b^', c, and c^', and let the polygon diagonals e, f, and g be so chosen that a, a^', and e have no common ...

Four or more points P_1, P_2, P_3, P_4, ... which lie on a circle C are said to be concyclic. Three points are trivially concyclic since three noncollinear points determine a ...

If P be a point in the plane of an equilateral triangle DeltaABC, then the lengths of line segments AP, BP, and CP correspond the sides of a triangle, which is degenerate ...

If the four points making up a quadrilateral are joined pairwise by six distinct lines, a figure known as a complete quadrangle results. A complete quadrangle is therefore a ...

The Tusi couple is a 2-cusped hypocycloid obtained by rolling a circle of radius a inside a circle of radius 2a. The result is a line segment (Steinhaus 1999, p. 145; Kanas ...

The regular pentagon is the regular polygon with five sides, as illustrated above. A number of distance relationships between vertices of the regular pentagon can be derived ...

In a given circle, find an isosceles triangle whose legs pass through two given points inside the circle. This can be restated as: from two points in the plane of a circle, ...