# Thread: Proof 2

1. ## Proof 2

Prove that if the consecutive mid-points of a rhombus be joined,a rectangle is formed.

2. This proof turns on a simple fact: The diagonals of a rhombus are perpendicular.
Then note that consecutive ‘midpoint segments’ are parallel to the diagonals.

3. How do I proceed? Can you please explain? I have proved without drawing the diagonals that the quadrilateral formed is a parallelogram, but i am unable to prove that it is a rectangle.Please help.
Thank you.

4. Originally Posted by muks
How do I proceed? Can you please explain? I have proved without drawing the diagonals that the quadrilateral formed is a parallelogram, but i am unable to prove that it is a rectangle.Please help.
Thank you.
I suggest using a vector proof. The dot product might prove useful .....

5. if you consider one half of the rhombus this will be a triangle with two of its sides joint by a line(name it L1) at their mid point. From congurence of the two triangles you can say that the base of the triangle is parallel to the line L1.
The same can be proved for the half rhombus that contains the other diagonal and let the line joining the mid point of the side be L2.
then you have L1 parallel to one of the rhombus diagonals and L2 parallel to the other diagonal. But the two diagonals of the rhombus are perpendicular then L1 and L2 are perpendicular. Then you have got a parallelogram with right angle (rectangle).