Here are two practice problems to test that you understand the proofs of Ceva and Menelaus.

Question 1.

In the barycentric coordinates relative to $\triangle ABC$ on which cevians does the point $G = \frac{2A + 3B + 5C}{10}$ lie?  Hint: Rewrite the expression for $G$ so that $G = xC + yC'$ where $C'$ lies on the opposite side from $C$ .

In effect, you need to find $C'$ for which $C'(AB)$ and $G(C C')$, for all three points . Here it is for $C'$ . Rewrite $G = ( \frac{2A+3B}{5} ) \frac{5}{10} + \frac{5}{10}C$ which has the form $G = C' \frac{5}{10} + \frac{5}{10}C$. Therefore $C' = (\frac{2A+3B}{5} )$ is the point you seek.

Question 2.

In the barycentric coordinates relative to $\triangle ABC$ let the coordinates be called $(a,b,c)$ for $aA+bB+cC$. Where does the line $2a + 3b + 5c =0$ cross the sides of the triangle? . Hint: We have that $a=0$ for all points on $(BC)$. And we solve $3b+5c=0$ for $b= -5$ and $c=3$. But $(0,-5,3)$ are not the barycentric coordinates of any point because they do not add to one, $0 \ne 0-5+3=-2 .$

The point $A' = \frac{5}{2}B - \frac{3}{2}C$ lies on $(BC)$ and on the line $2a+3b+5c=0$ and is therefore 1/3 of our answer.
As in Cartesian coordinates systems, the coordinates in barycentric coordinate stystems are unique, just as long as they add up to $1$. We can use this uniqueness to solve many geometrical problems.