Ad
related to: the diagonals of a rhombus are perpendicular proof worksheet
Search results
Results From The WOW.Com Content Network
In a cyclic orthodiagonal quadrilateral, the anticenter coincides with the point where the diagonals intersect. [3] Brahmagupta's theorem states that for a cyclic orthodiagonal quadrilateral, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side. [3]
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 ...
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]
This follows from the left side of the equation being equal to zero, requiring the right side to equal zero as well, and so the vector sum of a + b (the long diagonal of the rhombus) dotted with the vector difference a - b (the short diagonal of the rhombus) must equal zero, which indicates the diagonals are perpendicular.
Repeating this same argument with the other two points of tangency completes the proof of the result. If the extensions of opposite sides in a tangential quadrilateral intersect at J and K, and the diagonals intersect at P, then JK is perpendicular to the extension of IP where I is the incenter. [20]: Cor.4
Note 1: The most general trapezoids and isosceles trapezoids do not have perpendicular diagonals, but there are infinite numbers of (non-similar) trapezoids and isosceles trapezoids that do have perpendicular diagonals and are not any other named quadrilateral. Note 2: In a kite, one diagonal bisects the other.
The diagonals of a square are equal and bisect each other, meeting at 90°. The diagonal of a square bisects its internal angle, forming adjacent angles of 45°. All four sides of a square are equal. Opposite sides of a square are parallel. A square has Schläfli symbol {4}. A truncated square, t{4}, is an octagon, {8}.
In geometry, Brahmagupta's theorem states that if a cyclic quadrilateral is orthodiagonal (that is, has perpendicular diagonals), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side. [1] It is named after the Indian mathematician Brahmagupta (598-668). [2]