TITLE: Is every parallelogram a rectangle ??
QUESTION [31 upvotes]: 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?

REPLY [2 votes]: The answer by Narasimham is fully right.
To visualize it better, consider to flip $\triangle BCD$ around the bisector of the common segment $DC$.  

The resulting sketch shows that angles in $D$ (for $\triangle ADC$) and $C$ (for $\triangle BCD$) are supplementary.
They are equal only when they are rect, i.e. when the quadrilateral
is a rectangle.