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The area of a parallelogram is twice the area of a triangle created by one of its diagonals. The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. Any line through the midpoint of a parallelogram bisects the area. [6]
Given a norm, one can evaluate both sides of the parallelogram law above. A remarkable fact is that if the parallelogram law holds, then the norm must arise in the usual way from some inner product. In particular, it holds for the p {\displaystyle p} -norm if and only if p = 2 , {\displaystyle p=2,} the so-called Euclidean norm or standard norm.
From the first proof, one can see that the sum of the diagonals is equal to the perimeter of the parallelogram formed. Also, we can use vectors 1/2 the length of each side to first determine the area of the quadrilateral, and then to find areas of the four triangles divided by each side of the inner parallelogram.
A rhombus therefore has all of the properties of a parallelogram: for example, opposite sides are parallel; adjacent angles are supplementary; the two diagonals bisect one another; any line through the midpoint bisects the area; and the sum of the squares of the sides equals the sum of the squares of the diagonals (the parallelogram law).
Parallelogram law – Sum of the squares of all 4 sides of a parallelogram equals that of the 2 diagonals; Polarization identity – Formula relating the norm and the inner product in a inner product space; Ptolemy – Roman astronomer and geographer (c. 100–170) Ptolemy's table of chords – 2nd century AD trigonometric table
A parallelogram with equal diagonals is a rectangle. The Japanese theorem for cyclic quadrilaterals [ 12 ] states that the incentres of the four triangles determined by the vertices of a cyclic quadrilateral taken three at a time form a rectangle.
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.
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;