Parallelogram

Image:Parallelogram.jpg
A parallelogram.

A parallelogram is a four-sided plane figure that has two sets of opposite parallel sides. Every parallelogram is a polygon, and more specifically a quadrilateral. Special cases of a parallelogram are the rhombus, in which all four sides are of equal length, the rectangle, in which the two sets of opposing, parallel sides are perpendicular to each other, and the square, in which all four sides are of equal length and the two sets of opposing, parallel sides are perpendicular to each other. In any parallelogram, the diagonals bisect each other, i.e, they cut each other in half.

The parallelogram law distinguishes Hilbert spaces from other Banach spaces.

It is possible to create a tessellation with any parallelogram.

The three-dimensional counterpart of a parallelogram is a parallelepiped.

Proof that diagonals bisect each other

Prove that the diagonals of a parallelogram bisect each other. (Prove that E/AC=1:1 and E/BD=1:1)

(I'm not sure how to get vector symbols, so line segments such as AB are vectors. Feel free to fix the notation if you know how.) Proof: AE=kAC, k is an element of the real numbers

AE=k(AD+DC)

AE=k(AD+AB) since DC=AB

AE=kAD+kAB

since E,D,B are collinear, by the division-point theorem,

k+k=1

2k=1

k=0.5

sub k=0.5 into:

AE=kAC

AE=(0.5)AC

AE/AC=0.5 (the ratio of AE to AC is 1:2)

also sub k=0.5 into:

AE=kAD+kAB

AE=0.5AD+0.5AB

by the division-point theorem,

E/DB=1:1

by adding the division ratios to the parallelogram, we see that E divides both diagonals in the ratio 1:1, and E bisects AC and BD.

Therefore, the diagonals of a parallelogram bisect each other.


See also

External links

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