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The high school exterior angle theorem (HSEAT) says that the size of an exterior angle at a vertex of a triangle equals the sum of the sizes of the interior angles at the other two vertices of the triangle (remote interior angles). So, in the picture, the size of angle ACD equals the size of angle ABC plus the size of angle CAB.
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two interior angles that are not adjacent to it; this is the exterior angle theorem. [34] The sum of the measures of the three exterior angles (one for each vertex) of any triangle is 360 degrees, and indeed, this is true for any convex polygon, no matter ...
Exterior angles can be also defined, and the Euclidean triangle postulate can be formulated as the exterior angle theorem. One can also consider the sum of all three exterior angles, that equals to 360° [9] in the Euclidean case (as for any convex polygon), is less than 360° in the spherical case, and is greater than 360° in the hyperbolic case.
The formula can be proved by using mathematical induction: starting with a triangle, for which the angle sum is 180°, then replacing one side with two sides connected at another vertex, and so on. The sum of the external angles of any simple polygon, if only one of the two external angles is assumed at each vertex, is 2π radians (360°).
For the exterior angle bisectors in a non-equilateral triangle there exist similar equations for the ratios of the lengths of triangle sides. More precisely if the exterior angle bisector in A intersects the extended side BC in E , the exterior angle bisector in B intersects the extended side AC in D and the exterior angle bisector in C ...
Angle bisector theorem; ... Exterior angle theorem; H. Heron's formula; ... Midpoint theorem (triangle) Mollweide's formula; Morley's trisector theorem; N.
The exterior angle here could be called a supplementary exterior angle. Exterior angles are commonly used in Logo Turtle programs when drawing regular polygons. In a triangle, the bisectors of two exterior angles and the bisector of the other interior angle are concurrent (meet at a single point). [18]: 149
Therefore, triangle VOA is isosceles, so angle ∠BVA (the inscribed angle) and angle ∠VAO are equal; let each of them be denoted as ψ. Angles ∠BOA and ∠AOV are supplementary, summing to a straight angle (180°), so angle ∠AOV measures 180° − θ. The three angles of triangle VOA must sum to 180°: