Find the value of x for which ABCD must be a parallelogram.

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anonymous · Answer

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owlcoffee · Answer

A parallelogram has a very interesting property which is a little theorem (if you want the proof for you it, I can link you to my tutotial on verctorial spaces where I tackle this).

But something important to note is that a parallelogram has two diagonals and these two diagonals actually intersect eachother on their mid-points.
What does that imply?
Let's call the point of intersection of your parallelogram as "M" so this means that:
$$dist(A,M)=dist(M,C)$$
Ain't that a charm? and we have these distances, they are $2x+8$ and $6x$ so it's just a matter of plugging them in:
$$dist(A,M)=dist(M,C) \iff 2x+8=6x$$
And now, you have a first degree equality, which I have to suppose you can solve and thus obtaining the desired value of "x".