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Any two equilateral triangles are similar. Two triangles, both similar to a third triangle, are similar to each other (transitivity of similarity of triangles). Corresponding altitudes of similar triangles have the same ratio as the corresponding sides. Two right triangles are similar if the hypotenuse and one other side have lengths in the ...
The apparent triangles formed from the figures are 13 units wide and 5 units tall, so it appears that the area should be S = 13×5 / 2 = 32.5 units. However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent. With the bent ...
An arbitrary quadrilateral and its diagonals. Bases of similar triangles are parallel to the blue diagonal. Ditto for the red diagonal. The base pairs form a parallelogram with half the area of the quadrilateral, A q, as the sum of the areas of the four large triangles, A l is 2 A q (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, A s is a quarter of A ...
Figure 1: The point O is an external homothetic center for the two triangles. The size of each figure is proportional to its distance from the homothetic center. In geometry, a homothetic center (also called a center of similarity or a center of similitude) is a point from which at least two geometrically similar figures can be seen as a dilation or contraction of one another.
In geometry, Brahmagupta's theorem states that if a cyclic quadrilateral is orthodiagonal (that is, has perpendicular diagonals), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side. [1] It is named after the Indian mathematician Brahmagupta (598-668). [2]
In Euclidean geometry, the AA postulate states that two triangles are similar if they have two corresponding angles congruent. The AA postulate follows from the fact that the sum of the interior angles of a triangle is always equal to 180°. By knowing two angles, such as 32° and 64° degrees, we know that the next angle is 84°, because 180 ...
The circumcircles of all four triangles of a complete quadrilateral meet at a point M. [7] In the diagram above these are ∆ABF, ∆CDF, ∆ADE and ∆BCE. This result was announced, in two lines, by Jakob Steiner in the 1827/1828 issue of Gergonne's Annales de Mathématiques , [ 8 ] but a detailed proof was given by Miquel.
The Van Aubel points, the mid-points of the quadrilateral diagonals and the mid-points of the Van Aubel segments are concyclic. [3] A few extensions of the theorem, considering similar rectangles, similar rhombi and similar parallelograms constructed on the sides of the given quadrilateral, have been published on The Mathematical Gazette. [5] [6]
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