Measuring how similar to a parallelogram

[shybishie]

This is a (fairly basic) lemma without proof I saw in a research paper. Wasn't sure how to classify it exactly, but decided it's closest to vector (and linear) algebra.

It goes like this, consider a quadrilateral in the plane with vertices $$A$$, $$B$$, $$C$$, $$D$$ in clockwise order. It is given that $$||\vec{AB}|| = |\vec{CD}|| = 1$$.
Then show that $$||\vec{BC}||^2 -|| \vec{AD}||^2 \leq ||\vec{BA} - \vec{CD}||^2$$.

The intuition would be , $$\vec{BA} - \vec{CD}$$ measures how far away the quadrilateral $$ABCD$$ is from being a parallelogram, i.e, how much $$\vec{BC}$$ and $$\vec{AD}$$ differ in length. If anyone has some ideas or hints, it would be appreciated. I tried the law of sines, the law of cosines, and seeing if cross products would yield me any interesting relations - that didn't quite work out.

 

[AlephZero]

If you just use vector operations, I couldn't see a good way to represent the fact that the points are coplanar. So I invented a coordinate system with the origin at A and C on the x axis

A = (0,0)
B = (cos b, sin b)
C = (c, 0)
D = (c + dos d, sin d)

If required, we can specify the ranges of b and d so the points go clockwise round the quadrilateral.

I then get
|BC|^2 - |AD|^2 = -2c(cos b + cos d)

|BA - CD|^2 = 2 - 2 cos(b-d)

The first expression depends on the length of the diagonal c, but the second does not. Those two facts are geometrically obvious.

So the inequality can't be true for ANY value of c. Have you missed some part of the conditions?

If the length of c was irrelevant, the conditions could just specify |AB| = |CD|. It wouldn't need to say that both vectors have UNIT length.

 

[shybishie]

Ah, there is indeed a big typo on my part. The inequality is actually.

$$||\vec{BC}|| -|| \vec{AD}|| \leq ||\vec{BA} - \vec{CD}||$$.

And as the user Office_Shredder pointed out, this turns out to be fairly obvious.$$||\vec{BA} - \vec{CD}|| = ||[\vec{b}-\vec{a}] - [\vec{c}-\vec{d)]|| = || [\vec{b}-\vec{c}\] - [\vec{d}-\vec{a}] || \geq ||\vec{BC}|| - ||\vec{AD}||$$, with the last step by the reverse triangle inequality.

I apologize for any wasted time because of my error. Thank you for your time and assistance. The condition that the vectors have unit length actually turned out to be irrelevant , for this part of the proof that I extracted. Apologies.