help geometry

here is a representative image i made:

since both diagonals cot eachother in the middle, they form 4 triangles; by looking at the image, we can say that:

$ \overline{AO}=\frac{6}{2} \ \therefore \ \overline{CO}=\frac{6}{2}  $

$ \overline{DO}=\frac{10}{2} \ \therefore \ \overline{BO}=\frac{10}{2}  $

now there are two little rules for triangles 

1) sum of any two sides $>$ third side.

2) difference of any two sides $<$ third side. ( commonly used only to prove whether  the triangle exists or not )

lets consider the $\triangle BOC $ where side $\overline{BC}=$third side (the third and largest side of the triangle, but the smallest side of the parallelogram)

1)  $ \frac{10}{2}+\frac{6}{2} > \text{ third side} $

$ 5+3> \text{ third side} $

$ 8>  \text{ third side}  $

changing the sides, would also change the 'direction' of the less-than-sign, so logically, it would become larger-than-sign:

$   \text{ third side} <8 $

Hello Guest!

The greatest possible length of the shortest side of the parallelogram is greater than two by the smallest possible length.

  !