91影视

The objective is to prove that a parallelogram is a rhombus if and only if the diagonals are perpendicular.

In a rhombus all four sides are equal and the diagonals intersect at \(90\) degrees and in a parallelogram the opposite sides are equal and the diagonals bisect each other.

To prove that a parallelogram is a rhombus if and only if the diagonals are perpendicular. We suppose that parallelogram is a rhombus and we have to show that diagonals are perpendicular.

Let the parallelogram has two non-parallel vectors \(u\) and \(v\) and the diagonals of the parallelogram are \(u+v\) and \(u-v\).

Now, the dot product of \(u+v\) and \(u-v\) is

\(\left (u+v \right )\cdot \left ( u-v \right )=u\cdot u-u\cdot v+v\cdot u-v\cdot v\)

\(\left ( u+v \right )^{2}=u^{2}-v^{2}\)

\(\left ( u+v \right )=u^{2}-u^{2}\) (since parallelogram is a rhombus)

\(\left ( u+v \right )=0\)

Thus, the dot product of two vectors is zero.

Therefore, the vectors are orthogonal.

So, if a parallelogram is a rhombus the diagonals are perpendicular.

Now, let the diagonals of the parallelogram be perpendicular. We have to show that a parallelogram is a rhombus.

If diagonals are perpendicular. Then,

\(\left (u+v \right )\cdot \left ( u-v \right )=0\)

\(u^{2}-v^{2}=0\)

\(u=v\)

Thus, if the diagonals of a parallelogram are perpendicular then it is a rhombus.