Thread: analytic proof(s)

We are asked to show the analytic proof of the following:
"The diagonals of a parallelogram are equal in length. The figure is a rectangle."

>What I did was that I plotted a rectangle on a coordinate plane, and I used the distance formula to show that the diagonals are equal. The result was that the diagonals were indeed equal. But my question is this: "Can I already assume that what I've plotted was a rectangle (because it's obvious)? Or Should I still have to show that the figure is indeed a rectangle (through analytic proof) to be more precise?

thanks in advance..

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.

Dear teacher,
This is the actual sentence that was given to us:
"The diagonals of a parallelogram are equal length, the figure is a rectangle."

We are asked to prove that statement.

thanks in advance..