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Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. It follows that any rhombus has the following properties: Opposite angles of a rhombus have equal measure. The two diagonals of a rhombus are perpendicular; that is, a rhombus is an orthodiagonal quadrilateral. Its diagonals bisect opposite ...
An equidiagonal kite that maximizes the ratio of perimeter to diameter, inscribed in a Reuleaux triangle Among all quadrilaterals, the shape that has the greatest ratio of its perimeter to its diameter is an equidiagonal kite with angles π/3, 5π/12, 5π/6, and 5π/12.
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 ...
A rhombus with a right vertex angle; A rhombus with all angles equal; A parallelogram with one right vertex angle and two adjacent equal sides; A quadrilateral with four equal sides and four right angles; A quadrilateral where the diagonals are equal, and are the perpendicular bisectors of each other (i.e., a rhombus with equal diagonals)
The triangle ABC is a right triangle, as shown in the upper part of the diagram, with BC the hypotenuse. At the same time the triangle lengths are measured as shown, with the hypotenuse of length y, the side AC of length x and the side AB of length a, as seen in the lower diagram part. Diagram for differential proof
The solution to the quadrilateral isoperimetric problem is the square, and the solution to the triangle problem is the equilateral triangle. In general, the polygon with n sides having the largest area and a given perimeter is the regular polygon , which is closer to being a circle than is any irregular polygon with the same number of sides.
For a cyclic quadrilateral that is also orthodiagonal (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengths p 1 and p 2 and divides the other diagonal into segments of lengths q 1 and q 2. Then [28] (the first equality is Proposition 11 in Archimedes' Book of Lemmas)
The parallelogram between the pair of upright grey triangles has perpendicular diagonals in ratio , hence is a golden rhombus. If the triangle has legs of lengths 1 and 2 then each discrete spiral has length φ 2 = ∑ n = 0 ∞ φ − n . {\displaystyle \varphi ^{2}=\sum _{n=0}^{\infty }\varphi ^{-n}.}
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