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In Euclidean geometry, a parallelogram is a simple ... To prove that the diagonals of a parallelogram bisect each other, we will use congruent triangles:
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.
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 ...
The mathematical proof of the parallelogram of force is not generally accepted to be mathematically valid. Various proofs were developed (chiefly Duchayla's and Poisson's), and these also caused objections. That the parallelogram of force was true was not questioned, but why it was true. Today the parallelogram of force is accepted as an ...
The extended parallelogram sides DE and FG intersect at H. The line segment AH now "becomes" the side of the third parallelogram BCML attached to the triangle side BC, i.e., one constructs line segments BL and CM over BC, such that BL and CM are a parallel and equal in length to AH.
These various forms are all equivalent by the parallelogram law: [proof 1] ‖ ‖ + ‖ ‖ = ‖ + ‖ + ‖ ‖. This further implies that L p {\displaystyle L^{p}} class is not a Hilbert space whenever p ≠ 2 {\displaystyle p\neq 2} , as the parallelogram law is not satisfied.
As PGCH is a parallelogram, triangle PHE can be slid up to show that the altitudes sum to that of triangle ABC. This proof depends on the readily-proved proposition that the area of a triangle is half its base times its height—that is, half the product of one side with the altitude from that side.
The proof of the theorem is straightforward if one considers the areas of the main parallelogram and the two inner parallelograms around its diagonal: first, the difference between the main parallelogram and the two inner parallelograms is exactly equal to the combined area of the two complements;