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Not every parallelogram is a rhombus, though any parallelogram with perpendicular diagonals (the second property) is a rhombus. In general, any quadrilateral with perpendicular diagonals, one of which is a line of symmetry, is a kite. Every rhombus is a kite, and any quadrilateral that is both a kite and parallelogram is a rhombus. A rhombus is ...
Varignon's theorem holds that the midpoints of the sides of an arbitrary quadrilateral are the vertices of a parallelogram, called its Varignon parallelogram. If the quadrilateral is convex or concave (that is, not self-intersecting), then the area of the Varignon parallelogram is half the area of the quadrilateral. Proof without words (see ...
A rhombus is an orthodiagonal quadrilateral with two pairs of parallel sides (that is, an orthodiagonal quadrilateral that is also a parallelogram). A square is a limiting case of both a kite and a rhombus. Orthodiagonal quadrilaterals that are also equidiagonal quadrilaterals are called midsquare quadrilaterals. [2]
A Watt quadrilateral is a quadrilateral with a pair of opposite sides of equal length. [6] A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square. [7] A diametric quadrilateral is a cyclic quadrilateral having one of its sides as a diameter of the circumcircle. [8]
A convex quadrilateral is equidiagonal if and only if its Varignon parallelogram, the parallelogram formed by the midpoints of its sides, is a rhombus. An equivalent condition is that the bimedians of the quadrilateral (the diagonals of the Varignon parallelogram) are perpendicular. [3]
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 ...
To prove the quadrilateral case, simply construct the parallelogram tangent to the corners of the constructed rectangle, with sides parallel to the diagonals of the quadrilateral. The construction shows that the parallelogram is a rhombus, which is equivalent to showing that the sums of the radii of the incircles tangent to each diagonal are equal.
A tangential quadrilateral with two pairs of parallel sides is a rhombus. In this case, both midpoints and the center of the incircle coincide, and by definition, no Newton line exists. Proof