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The rhombus has a square as a special case, and is a special case of a kite and parallelogram. In plane Euclidean geometry, a rhombus (pl.: rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means
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 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.
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
The Varignon parallelogram is a rectangle if and only if the diagonals of the quadrilateral are perpendicular, that is, if the quadrilateral is an orthodiagonal quadrilateral. [6]: p. 14 [7]: p. 169 For a self-crossing quadrilateral, the Varignon parallelogram can degenerate to four collinear points, forming a line segment traversed twice.
Additionally, if a convex kite is not a rhombus, there is a circle outside the kite that is tangent to the extensions of the four sides; therefore, every convex kite that is not a rhombus is an ex-tangential quadrilateral. The convex kites that are not rhombi are exactly the quadrilaterals that are both tangential and ex-tangential. [16]
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 figure): An arbitrary quadrilateral and its diagonals. Bases of similar triangles are parallel to the blue diagonal. Ditto for the red diagonal.
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.