Centroids and area

There is a really surprising fact about centroids, one I had not known about until this year. The area of a triangle is equal to 1.5 times the distance of centroid to a side times the length of the side. It is not too hard to prove this result.

As I showed before, the area of a triangle is 0.5 (a*d - b*c) where u = (a, b) and v = (c, d) are two vectors that are the triangle's sides. To prove the really surprising fact about the centroid, all that must be shown is that $1.5·||g\perp v||·||v||=0.5·\left(ad-bc\right)$. The vector v is the side, so we need its magnitude; the perpendicular vector is adjusted by the scale factor g so that it is the proper distance from the side v to the centroid.

Finding g takes a little more work. Again as shown before, the vector w to the centroid is (u+v)/3, that is (a+c, b+d)/3. Using the Pythagorean theorem, it is true that ${||w||}^2={||g\perp v||}^2+{||fv||}^2$. One also knows from vector addition that $fv=w+g\perp v$. This allows a substitution for fv: . Therefore, substituting into the Pythagorean theorem yields:

This sets up showing that g = 0 (the trivial solution) or that:

$g=\frac{\left(a+c\right)·d-\left(b+d\right)c}{3\left(c^2+d^2\right)}=\frac{ad-bc}{3\left(c^2+d^2\right)}$

Therefore, our proof is nearly done. Going back to the original statement to prove,

$$\begin{gathered} 1.5·||g\perp v||·||v||=0.5·\left(ad-bc\right)\text{ \\ 1.5·g·c2+d2=0.5·ad-bc \\ 1.5·ad-bc3c2+d2·c2+d2=0.5·ad-bc} \end{gathered}$$

And our work is done.