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 $1.5\xb7||g\perp v||\xb7||v||=0.5\xb7\left(ad-bc\right)$. 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: $fv=>">a+c,b+d$. Therefore, substituting into the Pythagorean theorem yields:

${||w||}^{2}={||g\perp v||}^{2}+{||fv||}^{2}\phantom{\rule{0ex}{0ex}}2$

$g=\frac{\left(a+c\right)\xb7d-\left(b+d\right)c}{3\left({c}^{2}+{d}^{2}\right)}=\frac{ad-bc}{3\left({c}^{2}+{d}^{2}\right)}$

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

$1.5\xb7||g\perp v||\xb7||v||=0.5\xb7\left(ad-bc\right)\phantom{\rule{0ex}{0ex}}$

And our work is done.

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