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The pons asinorum in Oliver Byrne's edition of the Elements [1]. In geometry, the theorem that the angles opposite the equal sides of an isosceles triangle are themselves equal is known as the pons asinorum (/ ˈ p ɒ n z ˌ æ s ɪ ˈ n ɔːr ə m / PONZ ass-ih-NOR-əm), Latin for "bridge of asses", or more descriptively as the isosceles triangle theorem.
Since OA = OB = OC, OBA and OBC are isosceles triangles, and by the equality of the base angles of an isosceles triangle, ∠ OBC = ∠ OCB and ∠ OBA = ∠ OAB. Let α = ∠ BAO and β = ∠ OBC. The three internal angles of the ∆ABC triangle are α, (α + β), and β. Since the sum of the angles of a triangle is equal to 180°, we have
These include the Calabi triangle (a triangle with three congruent inscribed squares), [10] the golden triangle and golden gnomon (two isosceles triangles whose sides and base are in the golden ratio), [11] the 80-80-20 triangle appearing in the Langley's Adventitious Angles puzzle, [12] and the 30-30-120 triangle of the triakis triangular tiling.
Lexell's proof by breaking the triangle A ∗ B ∗ C into three isosceles triangles. The main idea in Lexell's c. 1777 geometric proof – also adopted by Eugène Catalan (1843), Robert Allardice (1883), Jacques Hadamard (1901), Antoine Gob (1922), and Hiroshi Maehara (1999) – is to split the triangle into three isosceles triangles with common apex at the circumcenter and then chase angles ...
An alternative proof (also based upon the triangle postulate) proceeds by considering three positions for point B: [10] (i) as depicted (which is to be proved), or (ii) B coincident with D (which would mean the isosceles triangle had two right angles as base angles plus the vertex angle γ, which would violate the triangle postulate), or lastly ...
Animated gif of proof of the inscribed angle theorem. The large triangle that is inscribed in the circle gets subdivided into three smaller triangles, all of which are isosceles because their upper two sides are radii of the circle. Inside each isosceles triangle the pair of base angles are equal to each other, and are half of 180° minus the ...
A direct proof using classical geometry was developed by James Mercer in 1923. [2] This solution involves drawing one additional line, and then making repeated use of the fact that the internal angles of a triangle add up to 180° to prove that several triangles drawn within the large triangle are all isosceles.
Set square shaped as 45° - 45° - 90° triangle The side lengths of a 45° - 45° - 90° triangle 45° - 45° - 90° right triangle of hypotenuse length 1.. In plane geometry, dividing a square along its diagonal results in two isosceles right triangles, each with one right angle (90°, π / 2 radians) and two other congruent angles each measuring half of a right angle (45°, or ...