How can I prove that under what condition, a parallelogram is (1) rectangle, (2)rhombus, (3) square?

The question is a follows.

Midpoints of the sides of a quadrilateral are the vertices of a paralleogram. Determine under what conditions this parallelogram will be (1) a rectangle, (2) a rhombus, (3) a square.

I tried this for my hw question but I am not sure if it was right. For (1), I think the two diagonals have to be perpendicular to each other, for (2), I wrote that the quadrilateral has to be rectangle. And for (3) I wrote that the quadrilateral has to be both (1) and (2). Can someone explain if this argument is right?? And if possible, can anyone give simple proof for this?

• You haven't provided any argument! Also, in a rhombus the diagonals are perpendicular - in a rectangle, the diagonals are of the same length. – user122283 Mar 4 '14 at 0:55
• @SanathDevalapurkar I meant the argument that I said the two diagonals have to be perpendicular to each other and etc.. I just wanted to know if these hypothesis is true. Thanks – eChung00 Mar 4 '14 at 0:59

1: The parallelogram is a rhombus if and only if the diagonals of the quadrilateral are perpendicular, that is, if the quadrilateral is an orthodiagonal quadrilateral

2: The parallelogram is a rectangle if and only if the two diagonals of the quadrilateral have equal length, that is, if the quadrilateral is an equidiagonal quadrilateral. Reference for those two theorems.

3 (Hint): When the quadrilateral is both a rectangle and a rhombus, then it is a square. So, ..

A parallelogram can be a

• Rectangle if, each of the angle is a right angle, diagonals are equal.

• Rhombus if, diagonals are perpendicular to each other.

• Square if each angle is 90 degree, diagonals are perpendicular and equal.