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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.
As another example, the solution set of a system of equations that depends on parameters generally has a fixed cardinality and dimension, but cardinality and/or dimension may be different for some exceptional values, called degenerate cases. In such a degenerate case, the solution set is said to be degenerate.
The midpoint polygon of a quadrilateral is a parallelogram called its Varignon parallelogram. If the quadrilateral is simple, the area of the parallelogram is one half the area of the original quadrilateral. The perimeter of the parallelogram equals the sum of the diagonals of the original quadrilateral.
Vectors involved in the parallelogram law. In a normed space, the statement of the parallelogram law is an equation relating norms: ‖ ‖ + ‖ ‖ = ‖ + ‖ + ‖ ‖,.. The parallelogram law is equivalent to the seemingly weaker statement: ‖ ‖ + ‖ ‖ ‖ + ‖ + ‖ ‖, because the reverse inequality can be obtained from it by substituting (+) for , and () for , and then simplifying.
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 l (half linear dimensions yields quarter area), and the area of the parallelogram is A ...
Splitting the thin parallelogram area (yellow) into little parts, and building a single unit square with them The key to the puzzle is the fact that neither of the 13×5 "triangles" is truly a triangle, nor would either truly be 13x5 if it were, because what appears to be the hypotenuse is bent.
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