Summary
In Euclidean geometry, an orthodiagonal quadrilateral is a quadrilateral in which the diagonals cross at right angles. In other words, it is a four-sided figure in which the line segments between non-adjacent vertices are orthogonal (perpendicular) to each other. A kite is an orthodiagonal quadrilateral in which one diagonal is a line of symmetry. The kites are exactly the orthodiagonal quadrilaterals that contain a circle tangent to all four of their sides; that is, the kites are the tangential orthodiagonal quadrilaterals. 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 equidiagonal quadrilaterals in which the diagonals are at least as long as all of the quadrilateral's sides have the maximum area for their diameter among all quadrilaterals, solving the n = 4 case of the biggest little polygon problem. The square is one such quadrilateral, but there are infinitely many others. An orthodiagonal quadrilateral that is also equidiagonal is a midsquare quadrilateral because its Varignon parallelogram is a square. Its area can be expressed purely in terms of its sides. For any orthodiagonal quadrilateral, the sum of the squares of two opposite sides equals that of the other two opposite sides: for successive sides a, b, c, and d, we have This follows from the Pythagorean theorem, by which either of these two sums of two squares can be expanded to equal the sum of the four squared distances from the quadrilateral's vertices to the point where the diagonals intersect. Conversely, any quadrilateral in which a2 + c2 = b2 + d2 must be orthodiagonal. This can be proved in a number of ways, including using the law of cosines, vectors, an indirect proof, and complex numbers. The diagonals of a convex quadrilateral are perpendicular if and only if the two bimedians have equal length.
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