Is quadrilateral ABCD a parallelogram ?

Statement 1:



Statement 2:



Statements 1 and 2 together:

This is where it gets really interesting. We know the diagonals are perpendicular, and the four sides are 8, 8, 6, and 6.

There are only two possible ways to arrange the sides. Either the opposite sides are equal, like this:



Or the adjacent sides are equal, like this:



However, if the opposite sides are equal, the two diagonals can't be perpendicular.

Here's the simplest argument I could come up with to demonstrate that:



The diagonals of a parallelogram should split each other in half. If they're perpendicular (have a right angle between them), we should be able to use right triangle rules! So, x^2 + y^2 = 6^2. However, we also have x^2 + y^2 = 8^2. It's impossible for the same equation to have two different values like that - so, it's logically impossible for the diagonals to be perpendicular.

In other words, if the opposite sides are equal, Statement 2 can't also be true. It's mathematically impossible. But Statement 2 has to be true, since we're combining the two statements together.

The only possible way to combine the two statements together and have them both be true, is if the adjacent sides are equal. This shape is the only possible one that fits both statements:



This shape is not a parallelogram. So, we can definitively answer 'no' to the question. The correct answer is C.