<s>[INST] 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?
 [/INST] As already hinted in comments and answers, you appear to have
applied Heron's formula
to the areas of the triangles $\triangle ADC$ and $\triangle BCD$.
Then observing that the only variables in Heron's formula are
the three sides of the triangle and its area,
and that three of these quantities in the formula for $\triangle ADC$
are equal to the corresponding quantities in the formula
for $\triangle BCD$, you concluded that the fourth quantity
must also be the same.
I will try to show that Heron's formula not only allows for multiple
solutions but (in the case of a non-rectangular parallelogram)
gives exactly the two distinct lengths of the diagonals.
One usually sees Heron's formula in a symmetric form such as this:
\begin{align}
s &= \tfrac12(a+b+c),\\
\triangle &= \sqrt{s(s-a)(s-b)(s-c)}.
\end{align}
This format is slightly deceptive since the second line shows
only one explicit occurrence of the variable $c,$ but there are
four implicit occurrences of $c$, one in each place where $s$ occurs. 
A more explicit formula is
$$
\triangle = \tfrac14 \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}.
$$
Algebra shows that the quantity inside the radical can be regrouped
as follows:
$$
\triangle = \tfrac14 \sqrt{(2ab)^2 - (a^2+b^2-c^2)^2}.  \tag1
$$
Now, given 
$\triangle = \text{Area}(\triangle ADC) = \text{Area}(\triangle BCD),$ 
$a = AD = BC,$ and $b = CD,$
suppose that $c$ is a solution to Equation $1.$
In order for the quantity under the radical in Equation $1$
to be positive, it must be true that $c^2\leq 2(a^2+b^2).$
Let
$$
c' = \sqrt{2(a^2+b^2) - c^2}.  \tag2
$$
Squaring both sides of Equation $2$ and rearranging terms,
$$
a^2 + b^2 - c'^2 = -(a^2+b^2-c^2)  \tag3
$$
and the squares of both sides of Equation $3$ are equal.
Hence we can substitute $(a^2 + b^2 - c'^2)^2$ for
$(a^2 + b^2 - c^2)^2$ in Equation $1$,
that is, $c'$ is also a solution of that equation.
It is still possible that $c'$ is the same as $c.$
This happens precisely when $c^2 = a^2 + b^2,$
that is, when $\triangle ADC$ and $\triangle BCD$ are right triangles.
In that case the parallelogram is a rectangle, its diagonals are equal,
and the two triangles are congruent.
But any value of $c$ such that $\lvert a - b \rvert < c < a+b$
may be a diagonal of a parallelogram with sides $a$ and $b,$
and for every such value such that $c \neq \sqrt{a^2 + b^2}$
there is a distinct value $c'$ that is the length of the other diagonal.
</s>