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Any non-self-crossing quadrilateral with exactly one axis of symmetry must be either an isosceles trapezoid or a kite. [5] However, if crossings are allowed, the set of symmetric quadrilaterals must be expanded to include also the crossed isosceles trapezoids, crossed quadrilaterals in which the crossed sides are of equal length and the other sides are parallel, and the antiparallelograms ...
The Gergonne triangle (of ) is defined by the three touchpoints of the incircle on the three sides. The touchpoint opposite A {\displaystyle A} is denoted T A {\displaystyle T_{A}} , etc. This Gergonne triangle, T A T B T C {\displaystyle \triangle T_{A}T_{B}T_{C}} , is also known as the contact triangle or intouch triangle of A B C ...
In 3-dimensions it will be a zig-zag skew icositetragon and can be seen in the vertices and side edges of a dodecagonal antiprism with the same D 12d, [2 +,24] symmetry, order 48. The dodecagrammic antiprism, s{2,24/5} and dodecagrammic crossed-antiprism, s{2,24/7} also have regular skew dodecagons.
The trapezius [4] is a large paired trapezoid-shaped surface muscle that extends longitudinally from the occipital bone to the lower thoracic vertebrae of the spine and laterally to the spine of the scapula. It moves the scapula and supports the arm. The trapezius has three functional parts: an upper (descending) part which supports the weight ...
Three sides (SSS) Two sides and the included angle (SAS, side-angle-side) Two sides and an angle not included between them (SSA), if the side length adjacent to the angle is shorter than the other side length. A side and the two angles adjacent to it (ASA) A side, the angle opposite to it and an angle adjacent to it (AAS).
The sides are in the ratio 1 : √ 3 : 2. The proof of this fact is clear using trigonometry. The geometric proof is: Draw an equilateral triangle ABC with side length 2 and with point D as the midpoint of segment BC. Draw an altitude line from A to D. Then ABD is a 30°–60°–90° triangle with hypotenuse of length 2, and base BD of length 1.
Let O be the center of a circle, as in the diagram at right. Choose two points on the circle, and call them V and A. Draw line OV and extended past O so that it intersects the circle at point B which is diametrically opposite the point V. Draw an angle whose vertex is point V and whose sides pass through points A, B. Draw line OA.
In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.