Hello, mamen!

You did it backwards.

You should have stated the theorem clearly

**If** the diagonals of a parallelogram are equal in length, **then** the figure is a rectangle.

First, graph a parallelogram.

Code:

| (r,h) (b+r,h)
| D o - - - - - - - - o C
| /: * * /
| / : * * /
| / :h * /
| / : * * /
|/ * : * /
A o - - + - - - - - o B - -
(0,0) r (b,0)

We have parallelogram *ABCD.*

The base is: .b = AB, .b ≠ 0

Vertex *D* is *(r, h)*, where *r* is the horizontal displacement of *D* and *h* is the height.

. . Then vertex *C* is *(b+r, h).*

Then: .AC^2 .= .(b + r)^2 + h^2

And: . BD^2 .= .(r - b)^2 + h^2

Equate: .(b + r)^2 + h^2 .= .(r - b)^2 + h^2

. b^2 + 2br + r^2 + h^2 .= .r^2 - 2br + b^2 + h^2

And we have: .4br = 0 . → . b = 0, r = 0

Since b ≠ 0, we have: r = 0.

That is, *D* is directly above *A*: .__/__ DAB = 90 degrees.

Therefore, *ABCD* is a rectangle.