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The longest diagonals of a regular hexagon, connecting diametrically opposite vertices, are twice the length of one side. From this it can be seen that a triangle with a vertex at the center of the regular hexagon and sharing one side with the hexagon is equilateral, and that the regular hexagon can be partitioned into six equilateral triangles.
Image mnemonic to help remember the ratios of sides of a right triangle. The sine, cosine, and tangent ratios in a right triangle can be remembered by representing them as strings of letters, for instance SOH-CAH-TOA in English: Sine = Opposite ÷ Hypotenuse Cosine = Adjacent ÷ Hypotenuse Tangent = Opposite ÷ Adjacent
Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine: sin(α + β) = sin α cos β + cos α sin β.
The parameters most commonly appearing in triangle inequalities are: the side lengths a, b, and c;; the semiperimeter s = (a + b + c) / 2 (half the perimeter p);; the angle measures A, B, and C of the angles of the vertices opposite the respective sides a, b, and c (with the vertices denoted with the same symbols as their angle measures);
Ptolemy's theorem is a relation among these lengths in a cyclic quadrilateral. = + In Euclidean geometry, Ptolemy's theorem is a relation between the four sides and two diagonals of a cyclic quadrilateral (a quadrilateral whose vertices lie on a common circle).
If an angle of a triangle is bisected internally or externally by a straight line which cuts the opposite side or the opposite side produced, the segments of that side will have the same ratio as the other sides of the triangle; and, if a side of a triangle be divided internally or externally so that its segments have the same ratio as the ...
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If the internal bisector of angle A in triangle ABC has length and if this bisector divides the side opposite A into segments of lengths m and n, then [3]: p.70 + = where b and c are the side lengths opposite vertices B and C; and the side opposite A is divided in the proportion b:c.