Is every parallelogram a rectangle ??

Let's say we have a parallelogram $\text{ABCD}$.

$\triangle \text{ADC}$ and $\triangle \text{BCD}$ are on the same base and between two parallel lines $\text{AB}$ and $\text{CD}$, So, $$ar\triangle \text{ADC}=ar\triangle \text{BCD}$$ Now the things those should be noticed are that:

In $\triangle \text{ADC}$ and $\triangle \text{BCD}$:

$$\text{AD}=\text{BC}$$ $$\text{DC}=\text{DC}$$ $$ar\triangle \text{ADC}=ar\triangle \text{BCD}.$$

Now in two different triangles, two sides are equal and their areas are also equal, so the third side is also equal or $\text{AC}=\text{BD}$. Which make this parallelogram a rectangle.

Isn't it a claim that every parallelogram is a rectangle or a parallelogram does not exist?

Your reasoning is appreciable, but the problem is that it is wrong.

While tempering with sides and area, you forgot about the angles. In this case , it is just the matter of sines and cosines. Let's see how:

Suppose $\angle ADC=\theta$ and $\angle BCD=180-\theta$.

On using trigonometric formula for area of a triangle, you will get that:

$$ar\triangle ADC=\frac{1}{2}\times AD\times DC\times \sin\theta$$

And , $$ar\triangle BCD=\frac{1}{2}\times BC\times CD \times \sin(180-\theta)=\frac{1}{2}\times BC\times CD \times \sin\theta$$

So, $ar\triangle ADC=ar\triangle BCD$

Now, using Cosine Formula:

$$AC^2=AD^2+DC^2-2\times AD\times DC\times \cos \theta$$

And, $$BD^2=BC^2+CD^2-2\times BC\times CD \times \cos (180-\theta)=BC^2+CD^2+2BC\times CD\times \cos\theta$$

This is enough to show that despite of having same area of $\triangle ADC$ and $\triangle BCD$, We can not say that $AC$ is equal to $BD$, these two are equal only when $\theta=90$ which is obviously the case of a rectangle.

EDIT:

After coming back on the site I saw that this question has got much attention and your comment made me realise that I should improve my post.

You discussed that you came to the result In two different triangles, two sides are equal and their Area is also equal. So, the third side is also equal. by Heron's Formula. Let's see when we approach this problem with Heron's Formula.

For area of a given triangle, We have $\triangle=\sqrt{s(s-a)(s-b)(s-c)}$ where terms have their usual meanings.

$$\triangle^2=s(s-a)(s-b)(s-c)=\frac{1}{16}(a+b+c)(a+b-c)(a+c-b)(c-a+b)$$

$$16\triangle^2=((a+b)^2-c^2)(c^2-(a-b)^2)$$

$$-16\triangle^2=c^4-2c^2\times (a^2+b^2)+(a^2-b^2)^2$$

$$c^4-2c^2\times (a^2+b^2)+[(a^2-b^2)^2+16\triangle^2]=0$$

Notice that this a polynomial of fourth degree and it is not necessary that all four values of $c$ are equal. The answer of your question comes exactly from here.

If you know two sides and the area of a triangle, there will generally be two different lengths for the third side that gives you that area.

Consider, for example: If the two known sides are $3$ and $4$, then the third side is somewhere between $1$ and $7$. A third side of length $1$ gives area $0$, but so does a third side of length $7$. In between, as the third side increases from $1$, the area of the triangle will first increase until it reaches a maximum of $6$ (when the third side is $5$), but then the area decreases towards $0$.

Thus every area between $0$ and $6$ will be hit by some third-side between $1$ and $5$, and by some third-side between $5$ and $7$.

If you try to use Heron's formula to derive the third side, you will end up with quadratic equation with two solutions.

Here is the problem:

"Two sides are equal and Area is also equal. So, the third side is also equal"

Take as an example:

Suppose $\angle ADC=60°$ and $\angle BCD =120°$.

You keep getting: $AD=BC$, $DC=DC$ and $S(ADC)=S(BCD)$ but $\Delta ADC \ne \Delta BCD$ because they don't fit the rule (side,angle,side):

$AD=BC$, $DC=DC$ but $\angle ADC \ne \angle BCD$.

P.S: Having the same area is a consequence of both triangle have the same base ($CD$) and the same height (because $AB$ is parallel to $CD$) but that doesn't garantee congruence. In order to see that just move the side $AB$ (keeping it parallel to $CD$ and on the same height) and clearly you will change the angles $\angle ADC$ and $\angle BCD$ but the area will keep the same.

EDIT

This post is getting some much attention that I think there is something more to say.

In the geometric construction above we have the line $t$ parallel to $s$. Those two paralelogram are just examples about what is happening. For any choice of $AB$ we will always have $AD=BC$ and $DC=DC$ and once $s$ and $t$ are parallels then the height $h$ is constant. It means that we will always have:

$$ar(ADC)=ar(BCD)=\frac{m\cdot h}{2}$$

and clearly there are infinite angles $\angle ADC$ that are not $90º$ and the parallelogram will not be a rectangle. Furthermore if $\angle ADC \ne 90º$ then $\angle ADC \ne \angle BCD$ and once the rule (side,angle,side) can be the definition of congruence between two triangles we will get $BD \ne AC$.