ভরকেন্দ্রিক স্থানাংকে ক্ষেত্রফল

Consider the diagram.

We have an arbitrary point G in the picture. Let's say the representation of G in bary is $Ax+By+Cz$ with $x+y+z=1$ of course.

Let's take a look at some easy cases first.

Now what is the ratio of [ABG]:[BGC]:[CGA]?

It's 1:1:1. This is well-known. Notice this also equals $z:x:y$

In fact, of any point P in the triangle has representation $Ax+By+Cz=P$ then we have

$[BCP]:[ACP]:[ABP] = x:y:z$ . This is not hard to prove: We can extend AP,BP,CP to meet the sides of the triangle.

Let the intersections of AP, BP, CP with the sides of the triangle be D, E, F, respectively.

Then clearly the ratio of $[APC]:[BPC] = [ACF]-[APF] : [BCF]-[BPF] = AF : BF$

This is because when two triangles have the same height, the ratio of the areas is the ratio of the bases.

We can do this for the other two ratios, and putting everything together gives you the theorem.

This should be straightforward.

Demonstration of barycentric coordinates essentials

Some people may wonder, how useful are these barycentric coordinates? Answer: Much more powerful than you think they are!

I just came across this article a few minutes before writing this, so allow me to demonstrate the power of barycentric coordinates!

Quoting from the above article:

Van Aubel's Theorem

As shown in the diagram, $ABC$ is a triangle with concurrent cevians $CD, BE, AF$. Then, van Aubel's Theorem states that $\frac{CE}{EA} + \frac{CF}{FB} = \frac{CG}{DG}$.

Proof by barycentric coordinates. Suppose we let $D=(p:q:0), E=(p:0:r), F=(0:q:r)$, and this is possible due to Ceva's theorem. If you follow this, it can be easily seen that $G=(p:q:r)$. Now $$\frac{CE}{EA} + \frac{CF}{FB} = \frac qr + \frac pr = \frac{(p+q+r)-(p+q)}{r} = \frac{CD-DG}{DG} = \frac{CG}{DG}$$ and we are done.