Search results
Results from the WOW.Com Content Network
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 ...
If the quadrilateral is a parallelogram, then the midpoints of the diagonals coincide so that the connecting line segment has length 0. In addition the parallel sides are of equal length, hence Euler's theorem reduces to + = + which is the parallelogram law.
In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
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.
Additionally, the four midpoints (grey) and the four feet of the maltitudes (red) are cocyclic on the 8-point-circle. A convex quadrilateral is orthodiagonal if and only if its Varignon parallelogram (whose vertices are the midpoints of its sides) is a rectangle. [6]
The topic of this article may not meet Wikipedia's general notability guideline. Please help to demonstrate the notability of the topic by citing reliable secondary sources that are independent of the topic and provide significant coverage of it beyond a mere trivial mention.
The same is true for BD, and so, ABD'C is a parallelogram. AD' is clearly the median, because a parallelogram's diagonals bisect each other, and AD is its reflection about the bisector. third proof. Let ω be the circle with center D passing through B and C, and let O be the circumcenter of ABC. Say lines AB, AC intersect ω at P, Q, respectively.
If the two sums of areas of opposite triangles are equal: | | + | | = | | + | |, then the point L is located on the Newton line, that is the line which connects E and F. [1] [2] For a parallelogram, the Newton line does not exist since both midpoints of the diagonals coincide with point of intersection of the diagonals.