Perhaps all parallelogram is also a rectangle?

No, a rectangle necessarily has angles of 90°.

Yes, a parallelogram is a type of rectangle.

No, a rectangle necessarily has angles of 90°.

From a Parallelogram to a Rectangle

From the Parallelogram to the Rectangle

Do you want to know how to prove that the parallelogram in front of you is actually a rectangle?
First, you should know that the formal definition of a rectangle is a parallelogram whose angle is $90^o$ degrees.
Additionally, if the diagonals in parallelograms are equal, it is a rectangle.

That is, if you are given a parallelogram, you can prove it is a rectangle using one of the following theorems:

If a parallelogram has an angle of $90^o$ degrees, it is a rectangle.
If the diagonals are equal in a parallelogram, it is a rectangle

We briefly remind you of the conditions for a parallelogram check:

If in a quadrilateral where each pair of opposite sides are also parallel to each other, the quadrilateral is a parallelogram.
If in a quadrilateral where each pair of opposite sides are also equal to each other, the quadrilateral is a parallelogram.
If a quadrilateral has a pair of opposite sides that are equal and parallel, the quadrilateral is a parallelogram.
If in a quadrilateral the diagonals cross each other, the quadrilateral is a parallelogram.
If a quadrilateral has two pairs of equal opposite angles, the quadrilateral is a parallelogram.

Detailed explanation

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Test yourself on from a parallelogram to a rectangle!

Is this parallelogram a rectangle

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Let's demonstrate that a parallelogram is a rectangle using the first theorem.

If a parallelogram has an angle of $90^o$ degrees, it is a rectangle.
We are given a parallelogram $ABCD$ :

From Parallelogram to Rectangle

Given that:
$∢B=90$

We need to prove that:
$ABCD$ is a rectangle.

Solution:

We know that in a parallelogram, the opposite angles are equal. Therefore, we can affirm that:
$∢B=∢D=90$
Now, we can affirm that:
$∢C=180-90$
Since the adjacent parallel angles are equal to $180^o$ .
We obtain that:
$∢C=90$

Wonderful. Now we can affirm that:
$∢A=∢C=90$
Since the opposite parallel angles are equal.

Magnificent! We proved that all angles in a parallelogram are equal to $90^o$ degrees.
Therefore, we can determine that the parallelogram is a rectangle.

Remember, the formal definition of a rectangle is a parallelogram where it has an angle of $90^o$ degrees.
Therefore, we will not have to prove that all angles are equal to $90^o$ .

Now, let's demonstrate that a parallelogram is a rectangle using the second theorem:

If the diagonals are equal in a parallelogram, it is a rectangle.
We are given a parallelogram $ABCD$ :

Parallelogram

and it has equal diagonals:
$AC=BD$

It is necessary to prove that: $ABCD$ is a rectangle.

Solution:
We know that in the parallelogram the diagonals cross each other.
Therefore, we can determine that:
$AE=CE$
$BE=DE$

We also know from the given data that: $AC=BD$
Therefore, we can affirm that all halves are equal.
Why?
We can write that:
$AE=CE=\frac{AC}{2}$

$BE=DE=\frac{BD}{2}$

Since:
$AC=BD$
we can compare
$\frac{AC}{2}=\frac{BD}{2}$
And according to the transitive rule we obtain:

$AE=CE=BE=DE$

As all these segments are equal, isosceles triangles are created.

We will identify the equal angles in the drawing:

Parallelogram

Opposite equal sides, the angles are equal.
Moreover, in a parallelogram, the opposite sides are parallel
and the alternate angles between parallel lines are equal.

Wonderful.
Now, remember that in a quadrilateral the sum of the interior angles equals $360^o$ .
Therefore:
$α+β+α+β+α+β+α+β=360$
$4α+4β=360$
we divide by $4$ and obtain:
$α+β=90$

Note that each of our parallel angles consists of $α+β$ and, therefore, each parallel angle equals $90^o$ degrees.
Therefore, the parallelogram we have in front of us is a rectangle, since a rectangle is a parallelogram with an angle of $90^o$ degrees.