Vector proofs

[monkeywantsabanana]

Using a vector method to prove if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

Thanks in advance.

[m@tty]

To make a parallelogram, you have two vectors, .

For it to also be a rhombus, the sides must have the same magnitude ()

The diagonals are and

If they are perpendicular, their dot product is zero:







And it is a rhombus, as defined above.

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

[monkeywantsabanana]

Thank you - again.

[monkeywantsabanana]

How do i also prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from all vertices?

[moekamo]

have the right angled corner as O, the others as A and B, the midpoint of hypotonuse is M.
vector OA is a, and vector BO is b.

Then BA=b+a
BM=0.5BA=0.5(b+a)
AM=-BM=-0.5(b+a)
OM=-OB+BM=0.5(a-b)

then you can show that these all have the same magnitude, done

note also that a is perpendicular to b, i.e a.b=0

[jasoN-]

        B         
       / |
      /  |
X   /\  |
    /  \ | 
   /__ \|
 O     A

Let OA = ai
Let AB = bj



To prove that the midpoint of the hypotenuse is equidistant from all vertices |OX| = |XB| = |AX|




edit: correct me if im wrong lol

[moekamo]

        B         
       / |
      /  |
X   /\  |
    /  \ | 
   /__ \|
 O     A

Let OA = a
Let AB = b



To prove that the midpoint of the hypotenuse is equidistant from all vertices |OX| = |XB| = |AX|

huh? you've somehow proven that the distance from the right angled corner to the midpoint of the hypotenuse is always sqrt(2)/2 !?!?!?

you cant just let a and be be 1(although you can make them in the i and j directions), for this to be general, they have to be any side lengths of any right angled triangle, you have to solve the magnitudes in terms of a and b.

[jasoN-]

oh sorry, you're right. im thinking complex numbers (like i and j) lol.
ill edit

[monkeywantsabanana]

Thanks guys !

[enpassant]

Here is a simpler proof.
right-angle triangle OAB, M is the midpoint of AB the hypotenuse, vector OA=a, vector OB=b
vector OM=1/2 (a+b)
(OM)^2=1/2 (a+b) dot 1/2 (a+b)
=1/4 (a+b) dot (a+b)
=1/4 (a^2+b^2) .......... because a and b are perpendicular, a dot b =0
=1/4 (AB)^2
.: OM=1/2 AB, end of proof